A 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle as seen in the figure.
If the ball is in contact with the wall for 0.207 s, what is the average force exerted on the ball by the wall?

A 3.42 Kg Steel Ball Strikes A Massive Wall At 14.3 M/s At An Angle Of = 61.5 With Respect To The Plane

Answers

Answer 1

The magnitude of the average force exerted on the ball by the wall is 225.469 N.

What is impulse?

The change in momentum is equal to the product of impact force applied while colliding and time for that impact.

Impulse F. t = m (Vf -Vi)

where, Vf is the final velocity and Vi is the initial velocity.

Given is a 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle. The ball is in contact with the wall for 0.207 s

Substitute the values into the expression, we get

Impulse = 2 x mvcosθ

Impulse= 2 x 3.42 x 14.3 x cos 61.5°

Impulse = 46.672 kg.m/s

The impact force can be written as

F.t = I

Put the given values, we have

F = 46.672 /  0.207

F = 225.469 N

Thus, the magnitude of force exerted by the wall is 225.469 N

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Related Questions

Part A
Calculate the spring constant.
• Look at the graph of x vs. t graph. (You may want to double-click on
it to examine it in the Data Tool view.)
. From the graph, determine the period T.
• The first video frame lists the mass.
• Use the equation below to solve for the spring constant k. Show
your work below.
m
T = 2π√√T

Answers

Answer:

Using 3 periods to get an accurate reading:

3T = (6.40 S - 0.90 s) = 5.50 S So, T = 1.83 s

m = 0.250 kg

Using algebra:

k= [tex]\frac{4\pi ^{2}m }{T^{2} }[/tex]=[tex]\frac{4\pi ^{2}0.250 }{1.90^{2} }[/tex]=2.95kg/sec

11. Trait theory claims that
O A. people from the same locations
share the same personality type.
B. you always behave the way your
personality type says you will.
O C. your personality is made up of a
number of traits.
O D. you have one characteristic that
defines your entire personality.

Answers

A. people from the same location share the same personality type.

Taking the density of air to be 1.29 kg/m3, what is the magnitude of the angular momentum (in kg · m2/s) of a cubic meter of air moving with a wind speed of 73.0 mi/h in a hurricane? Assume the air is 51.2 km from the center of the hurricane "eye."

Answers

The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.

Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h

We have to find the magnitude of the angular momentum

Let,

ρ = Density of air = 1.29 kg/m^3

v = Speed of wind = 73.0 mi/h = 0.032 km/s

M = angular momentum of air

Let the volume of air be 1 m^3

Mass = Volume x ρ = 1 x 1.29 = 1.29 kg

Momentum = M = mass x velocity

Momentum = 1.29 x 0.0032

Momentum = 4.128 x 10^(-3) kg·m^2/s

Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

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n Step 5, you will calculate H+/OH– ratios for more extreme pH solutions. Find the concentration of H+ ions to OH– ions listed in Table B of your Student Guide for a solution at a pH = 2. Then divide the H+ concentration by the OH– concentration. Record these concentrations and ratio in Table C.

What is the concentration of H+ ions at a pH = 2?

mol/L

What is the concentration of OH– ions at a pH = 2?

mol/L

What is the ratio of H+ ions to OH– ions at a pH = 2?

: 1

Answers

At pH = 2 the ratio of the concentration of the hydrogen to the hydroxide ion is 10¹⁰. The hydrogen ion concentration at pH, 2 is 10⁻² and the hydroxide ion is 10⁻¹².

What is pH?

The parameter of measuring the concentration of the hydroxide and the hydrogen ion in the solution in order to determine the basic and the acidic nature is known as pH.

The concentration of H⁺ ions is calculated as:

pH = -log [H⁺]

2 = -log [H⁺]

[H⁺] = 10⁻²

The concentration of OH⁻ ions is calculated as:

pH + pOH = 14

pOH = 14 - 2

pOH = 12

Solving further:

pOH = -log [OH⁻]

12 =-log [OH⁻]

[OH⁻] = 10⁻¹²

At a pH = 2, the ratio of hydrogen ions to hydroxide ions:

10⁻² ÷ 10⁻¹² = 10¹⁰

Therefore, at pH = 2 the ratio is 10¹⁰.

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A spring is compressed by 0.0880 m and is used to launch an object horizontally with a speed of 2.76 m/s. If the object were attached
to the spring, at what angular frequency (in rad/s) would it oscillate?

Answers

Answer:

Approximately [tex]3.14\; {\rm rad \cdot s^{-1}}[/tex].

Explanation:

Fact: the angular velocity [tex]\omega[/tex] of a simple harmonic oscillator is the ratio between the maximum velocity [tex]v_{\text{max}}[/tex] and the maximum displacement [tex]x_\text{max}[/tex] of this oscillator. In other words:

[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}}\end{aligned}[/tex].

Derivation of the previous equation:

Let [tex]A[/tex] denote the amplitude of this oscillation, and let [tex]\omega[/tex] denote the angular velocity.

The displacement of the oscillator at time [tex]t[/tex] would be:

[tex]x(t) = A\, \sin(\omega\, t)[/tex].

The maximum displacement of this oscillator would be [tex]x_\text{max} = A[/tex].

The velocity of this oscillator at time [tex]t[/tex] is the derivative of displacement with respect to time:

[tex]\begin{aligned} v(t) &= \frac{d}{d t}\, [x(t)] \\ &= \frac{d}{d t} [A\, \sin(\omega\, t)] \\ &= A\, \omega\, \cos(\omega\, t)\end{aligned}[/tex].

The maximum velocity of this oscillator would be [tex]v_\text{max} = A\, \omega[/tex].

Notice that dividing [tex]v_\text{max} = A\, \omega[/tex] by [tex]x_\text{max} = A[/tex] would give:

[tex]\displaystyle \frac{v_\text{max}}{x_\text{max}} = \frac{A\, \omega}{A} = \omega[/tex].

It is given that [tex]v_\text{max} = 2.76\; {\rm m\cdot s^{-1}}[/tex] while [tex]x_\text{max} = 0.0880\; {\rm m}[/tex]. Therefore:

[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}} \\ &= \frac{2.76\; {\rm m\cdot s^{-1}}}{0.0880\; {\rm m}} \\ &\approx 3.14\; {\rm s^{-1}}\end{aligned}[/tex].

(Radians per second.)

The action force is the ballon pushing the air out. What is the magnitude of the reaction force of the air pushing the ballon

Answers

The magnitude of the reaction force of the air pushing the balloon would be equal and opposite.

How is the reaction force equal and opposite?

A push or a pull that an object experiences as a result of interacting with another item is known as a  reaction force.

Newton's third law is officially expressed as follows: There is an equal and opposite reaction to every action. The implication of the statement is that there are always two forces acting on the two interacting objects. The force acting on the first object is equal in size to the force acting on the second. The force acting on the first object is acting in the opposite direction to the force acting on the second object. Force pairs—equal and opposing action-reaction force pairs—always exist in pairs.

Knowing that everything has an equal and opposite reaction according to Newton's second law. The preasured air that a balloon had to push out into the free air acts as the reaction force.

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Resolve the vector shown below into its components.

Answers

Answer:

D. s=3x+4y

Explanation:

The line is at the 3rd column in the 4th row.

During World War II, it was found that r, the radius of the shockwave produced during an atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density, .

Answers

The atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

What is an atomic bomb?

An atomic bomb is type of nuclear weapon that releases a vast amount of energy upon explosion in the form of fission reaction.

During an explosion of atomic bomb, a cloud of mushroom fire is formed which vaporises anything found within it. The extent of destruction depends on:

energy released, W,

the elapsed time, t, and

the air density.

Therefore, the atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

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Calculate the magnitude of the linear momentum for the following cases.
(a) a proton with mass 1.67 x 10-27 kg, moving with a speed of 5.05 x 106 m/s

(d) the Earth (mass - 5.98 x 1024 kg) moving with an orbital speed equal to 2.98 x 10 m/s.

Answers

Answer:

8.4335 x 10^-²¹, 1.78204 x 10²⁶

Explanation:

p = mv

p = (5.05 x 10⁶)(1.67 x 10^-²⁷)

p = 8.4335 x 10^-²¹

p = mv

p = (5.98 x 10²⁴)(2.98 x 10)

p = 1.78204 x 10²⁶

(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord can withstand a maximum tension of 64.0 N, a. What is the maximum speed at which the ball can move before the cord breaks? Assume the string remains horizontal during the motion. (5 marks) b. Suppose the ball moves in a circle of larger radius at the same speed v. Is the cord more likely or less likely to break? ( i ) . A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long . The ball moves in a horizontal circle . If the cord can withstand a maximum tension of 64.0 N , a . What is the maximum speed at which the ball can move before the cord breaks ? Assume the string remains horizontal during the motion . ( 5 marks ) b . Suppose the ball moves in a circle of larger radius at the same speed v . Is the cord more likely or less likely to break ?​

Answers

(a) Let [tex]v[/tex] be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

[tex]a_{\rm rad} = \dfrac{v^2}R[/tex]

where [tex]R[/tex] is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

[tex]F = (1.500\,\mathrm{kg}) a_{\rm rad}[/tex]

so that

[tex](1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N[/tex]

Solve for [tex]v[/tex] :

[tex]v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}[/tex]

(b) The net force equation in part (a) leads us to the relation

[tex]F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}[/tex]

so that [tex]v[/tex] is directly proportional to the square root of [tex]R[/tex]. As the radius [tex]R[/tex] increases, the maximum linear speed [tex]v[/tex] will also increase, so the cord is less likely to break if we keep up the same speed.

Determine the speed at which the medicine leaves the needle

Answers

The speed at which the medicine leaves the needle is  2.462 m/s

What is Bernoulli's theorem?

When an incompressible, ideal fluid is flowing through a tube or pipe, the total energy remains constant.

p₁ /ρg + v₁²/2g +z₁ = p₂ /ρg + v₂²/2g +z₂

Where, p/ρg = pressure energy

            v²/2g = kinetic energy

                   z = potential energy

Given is during an injection, pressure in the barrel of syringe is 1.03  atm while pressure in the needle section is 1.00 atm. Assuming the syringe lays horizontally and mass density of the liquid medicine ρ =1000 kg/m³.

The fluid is initially at rest.

Using the Bernoulli's equation, we have

v₁² -  v₂²= 2 x (p₂ -p₁) / ρ

Substituting the values, we get

0 -  v₂² = 2 x (1.00 -1.03) x 1.01 x 10⁵ /1000

v₂² = 6.06

v₂ = 2.462 m/s

Thus, the speed at which the medicine leaves the needle is 2.462 m/s

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A small sphere of mass 10 kg
is released from rest at a height of
15.0 m above the ground level.
The sphere experiences a constant
resistive force (due to air
resistance) of magnitude R = 10.0
N.
a) Calculate the speed of the
sphere after it has fallen
through a distance of 5.00 m

bCalculate the speed of the ball just before a it hits the gound.

Answers

Answer:

Approximately [tex]9.39 \; {\rm m\cdot s^{-1}}[/tex] after the sphere has travelled a distance of [tex]5\; {\rm m}[/tex].

Approximately [tex]16.3\; {\rm m\cdot s^{-1}}[/tex] right before touching the ground (a distance of [tex]15\; {\rm m}[/tex].)

Assumption: [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].

Explanation:

Weight of the sphere: [tex]m\, g = 9.81\; {\rm N \cdot kg^{-1}} \times 10\; {\rm kg} = 98.1\; {\rm N}[/tex], downwards.

Drag on the sphere: [tex]10.0\; {\rm N}[/tex] upwards.

Net force on the sphere: [tex]98.1\; {\rm N} - 10\; {\rm N} = 88.1\; {\rm N}[/tex] downwards.

Acceleration of the sphere: [tex]a = F_\text{net} / m = 88.1\; {\rm N} / (10\; {\rm kg}) = 8.81\; {\rm m\cdot s^{-2}}[/tex].

Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity ([tex]0[/tex] in this case, as the sphere was released from rest,) and [tex]x[/tex] is the distance (displacement) that the sphere has travelled so far.

Rearrange this equation to obtain an expression for [tex]v[/tex]:

[tex]\displaystyle v = \sqrt{2\, a\, x + u^{2}}[/tex].

For example, after the ball travelled a distance of [tex]5.00\; {\rm m}[/tex], [tex]x = 5.00 \; {\rm m}[/tex]:

[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 5.0\; {\rm m} + 0} \\ &\approx 9.39\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Similarly, [tex]x = 15.0\; {\rm m}[/tex] right before landing, such that:

[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 15.0\; {\rm m} + 0} \\ &\approx 16.3\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

What must the charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 600 N/C ? Use 9.80 m/s2 for the magnitude of the free-fall acceleration.

Answers

[tex]q = -21 * 10^{-6} C[/tex]

What is Free-fall acceleration?

The acceleration is constant and equal to the gravitational acceleration g which is 9.8 m/s at sea level on the Earth.

As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force,

mg =qE

[tex]1.45 * 10^{-3}(9.80)=q(600)\\\\q = 21 *10^{6} C\\\\[/tex]

and its sign must be negative so that it will have upward electric force

so it is

[tex]q = -21 * 10^{-6} C[/tex]

The charge of a particle of mass is [tex]-21 * 10^{-6} C[/tex]

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5. A uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring). (1) M Before Wi (II) wf V Ө ə Ө After Ө wf H ? (a) What kind of frictional force acts on the ring upon contact with the surface? (b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), find the coefficient of friction corresponding to the frictional force you mentioned in (a). (c) What is the increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly? The ring then rolls smoothly up a ramp of 0 = π/6 rad and H = 5 m [see figure (II)] (d) What is the horizontal distance, from the end of the ramp, at which the ring lands?

Answers

(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), the coefficient of friction corresponding to the frictional force is 0.612

(c) The increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands is 7.78 m

What is frictional force?

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.

Given is a uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring).

(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) angular frequency = velocity / radius

ωf = 10/2 = 5 rad/s and ωi = 20 rad/s

Angular acceleration, α = ωf - ωi /t

Put the values, we get

α = -15/5 = -3 rad/s²

Coefficient of friction, μ = a/g = rα/g

Plug the values, we get

μ = 0.612

Thus, the coefficient of friction corresponding to the frictional force is 0.612.

(c) The energy lost = heat generated

 energy lost = 1/2 Iω² + 1/2 Mv²

 energy lost = 1/2 MR²ω² + 1/2 Mv²

Plug the values, we get

 energy lost = 7500 J

Thus, the  increase in the thermal energy is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands

s= (v- 2gH) sin2θ /g

s = (10 - 2x (-9.81)x5 ) sin (2x π/6) / 9.81

s = 7.78m

Thus, the horizontal distance is  7.78 m

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differentiate e^x - 8x +7
please answer the questions asap​

Answers

Answer:

eˣ-8.

Explanation:

(eˣ-8x+7)'=eˣ-8.

An arrow is shot horizontally from the top of a building and it lands 200m from the foot of the building after 10s.Assuming air resistance is negligible,calculate the initial velocity of the arrow and the height of the building?

Need an answer urgently please

Answers

The initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.

What is velocity?

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is;

u is the initial velocity of fall = ?  m/sec

h is the distance of fall = 200 m

g is the acceleration of free fall = 9.81 m/sec²

H is the height of the building

t is the time period = 10 second

According to Newton's second equation of motion,

[tex]\rm h= ut+\frac{1}{2} gt^2 \\\\\ h-\frac{1}{2} gt^2 =ut \\\\ 200 - 0.5 \times(9.81) \times 10^2 = 10 u \\\\ u = - 29.05 \ m/sec[/tex]

- ve shows the direction is downward.The magnitude of the initial velocity is found as;

u = 29.05 m/sec

The height of the building

[tex]\rm H= ut+\frac{1}{2} gt^2 \\\\\ H = 29.05 \times 10 + 0.5 \times 9.81 \times 10^2 \\\\ H = 781 \ m[/tex]

Hence the initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.

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6. A metal block increases in temperature from 15 °C to 60 °C when supplied with 13 500 J of heat energy.
a) Calculate the heat capacity of the metal.
b) Calculate the specific heat capacity of the metal, if this sample has a mass of 0.75 kg.​

Answers

Answer:

Since 1 cal = 4.19 J     the heat Q received by the metal is

Q = 13,500 J * 1 C / 4.19 J = 3,222 calories

S = ΔQ / (ΔT * ΔM)  = 3222 cal / (45 deg C / 750 gm)

S = .095 cal / gm deg C

Note that specific heat capacity for Cu is .093 cal / gm deg C

What is the wavelength of a compression wave with a speed of 500 m/s and a frequency of 250 Hz?
Use this formula: λ = s/f
Select one:
a. 2 m
b. 1.25 m
c. 250 m
d. 750 m

Answers

Answer:

a (2m)

Explanation:

according to wavelenght = speed / frequency,

the wavelength is 2 m

a car is traveling along a straight road at a velocity of 30m/s when its engine cuts off. For the next ten seconds, the car slows down, and its average acceleration is a1. For the next five seconds, the car slows down further at a velocity of 24m/s, and its average acceleration is a2. The ratio of the average acceleration values is a1/a2=1.5. Find the velocity of the car at the end of the initial ten-second interval.​

Answers

Answer:

See below....the question is unclear....pick the answer you think they want

Explanation:

Car goes from 30 m/s    to   24 m/s   in 10 seconds

acceleration = change in velocity/change in time

                     =   -6 m/s   / 10 s   =  - 3/5   m/s^2  

The wording of the question is not very good.....perhaps they meant to say that after the 5 seconds the velocity is now 24 m/s ?

a1 *10   +   a2 *5   = -6 m/s  

            ( 6 m/s is the change in velocity from 30 to 24 m/s)    

    a1/a2 = 1.5    so   a2 = a1/1.5   (substitute this in)

10 a1   + 5 ( a1/1.5)  =-6

15 a1 + 5 a1  = -9

a1 =  - .45 m/s^2

The velocity of the car at the end of the initial ten-second interval, given that the car further slows down for the next five seconds, is 25.5 m/s

How to determine the velocity of the car at the end of the 10 s interval?

From the first statement, we have:

Initial velocity (u₁) = 30 m/sTime (t₁) = 10 secondsVelocity at the end of 10 s (v₁) = ?

v₁ = u₁ + a₁t₁

v₁ = 30 + (a₁ × 10)

v₁ = 30 + 10a₁ ......(1)

From the second statement, we have:

Time (t₂) = 5 sFinal velocity (v₂) = 24 m/sRation of acceleration (a₁/a₂) = 1.5Initial velocity (u₂) = Velocity at the end of 10 s

v₂ = u₂ + a₂t₂

24 = u₂ + (a₂ × 5)

24 = u₂ + 5a₂   .......(2)

But,

u₂ = v₁ = 30 + 10a₁

Thus, we have:

24 = 30 + 10a₁ + 5a₂

But,

a₁/a₂ = 1.5

a₁ = 1.5a₂ ..... (3)

Thus, we have:

24 = 30 + 10a₁ + 5a₂

24 = 30 + 10(1.5a₂) + 5a₂

24 = 30 + 15a₂ + 5a₂

24 = 30 + 20a₂

Collect like terms

24 - 30 = 20a₂

-6 = 20a₂

Divide both sides by 20

a₂ = -6 / 20

= -0.3 m/s²

Substituting the value of a₂ into equation 3, we have

a₁ = 1.5a₂

= 1.5 × -0.3

= -0.45 m/s²

Substitute the value of a₁ into equation 1 to obtain the velocity, v₁ at the end of the initial to 10 s

v₁ = 30 + 10a₁

= 30 + (10 × -0.45)

= 30 - 4.5

= 25.5 m/s

Thus, velocity of the car at the end of the initial ten-second interval is 25.5 m/s

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Which graph shows constant acceleration with correctly placed independent and dependent variables?

Answers

Answer:

a = (V2 - V1) / t      equation  for constant acceleration

if t the independent variable is placed on the abcissa or x-axis and V the dependent variable is placed on the ordinate or y-axis then the acceleration would be the slope at the point considered.

What is the energy of a photon of blue light with a frequency of 6.53 × 10¹4 Hz?

Answers

Answer:

V = 6.53 × 10^14

h = 6.626 × 10^-34

E = hV

E = (6.626 × 10^-34) (6.53 × 10^14)

E = 4.33 × 10^-19

A bike accelerates uniformly from rest to a speed of 10 m/s over a distance of 50 m. (a) Determine the acceleration of the bike. (b) how long will take to do that?​

Answers

Answer:

a) 0.2m/s

b) 5m/s

Explanation:

a) acceleration=∆v/∆t

v=10

t=50

10/50=1/5

=0.2m/s

b) time= d/s

d=50m

s=10

50/10

=5m/s

Answer:

See below

Explanation:

Average speed =  (0+ 10)/2 = 5 m/s

  then to cover 50 m     will take    50 m / 5 m/s  = 10 seconds

      change in velocity/ change in time = acceleration = 10/10 = 1 m/s^2

A projectile is fired at an upward angle of 45.0º from the top of a 265-m cliff with a speed of .sm 185 What will be its speed when it strikes the ground below? (Use conservation of energy.)

Answers

The final velocity of the projectile when it strikes the ground below is 198.51 m/s.

Time of motion of the projectile

The time taken for the projectile to fall to the ground is calculated as follows;

h = vt + ¹/₂gt²

where;

h is height of the cliffv is velocityt is time of motion

265 = (185 x sin45)t + (0.5)(9.8)t²

265 = 130.8t + 4.9t²

4.9t² + 130.8t - 265 = 0

solve the quadratic equation using formula method,

t = 1.89 s

Final velocity of the projectile

vyf = vyi + gt

where;

vyf is the final vertical velocityvyi is initial vertical velocity

vyf = (185 x sin45) + (9.8 x 1.89)

vyf = 149.322 m/s

vxf = vxi

where;

vxf is the final horizontal velocityvxi is the initial horizontal velocity

vxf = 185 x cos(45)

vxf = 130.8 m/s

vf = √(vyf² + vxf²)

where;

vf is the speed of the projectile when it strikes the ground below

vf = √(149.322²  +  130.8²)

vf = 198.51 m/s

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Which of the following statements is false?

A Heat is a form of energy while temperature is a measure of degree of hotness

B Heart and temperature are linearly related

C Heat can be measured by physical changes due to change in temperature

D Temperature can be measured by physical changes due to change in heat​

Answers

d. is wrong...the temperature cannot be measured due to physical changes

Using Hooke's law, what happens when the displacement is tripled?

Answers

If the displacement is tripled then the spring force also is tripled.

What is Hooke's law?

The amount of force required to stretch or compress the spring is known as the spring force. Its unit is Newton(N).

Froom the hooks law;

F = kx

Where,

F is the applied force

x is the displacement in case 1

Case 2;

x'=3x

F'=kx'

F'=3KX

F'=3x'

Where,

x' is the displacement when the force is tripled

F' is the force in the case2

Hence, if the displacement is tripled then the spring force also be tripled.

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The value of acceleration due to gravity (g) on Pluto is about 0.61 meters/second2. How much will an object that weighs 250 newtons on Earth weigh on Pluto? Note that the value of acceleration due to gravity on Earth is 9.8 meters/second2.

Answers

The weight of an object on pluto will be 15.56. Mass multiplied by the gravitational acceleration gives weight.

What is gravitation?

Gravitation is a natural law by which all things with all matter are attracted towards one another. Gravity is responsible for large-scale structures present in the Universe.

By dividing the object's weight on Earth by 9.8 m/s², as illustrated below, one may calculate the object's mass.

m = 250 N / 9.8 m/s²

m = 25.51 kg

Multiply the acquired mass by Pluto's gravitational acceleration (g);

W = (25.51 kg) x (0.61 m/s²)  

W = 15.56 N

Thus, the item will only be 15.56 N in weight.

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Answer: B.

15.6 newtons

Explanation: edmentum

A force of 20N is applied to the end of a wire of length 5m to produce an extension of 0.20mm
calculate: the stress on the wire.​

Answers

The stress on the wire is determined as  6.37 x 10⁶ N/m².

Stress on the wire

The stress on the wire is calculated as follows;

σ = F/A

where;

F is the force applied on the wireA is area of the wire

Let the diameter of the wire = 2 mm

Area = πr²

Area = π(1 x 10⁻³)²

Area = 3.142 x 10⁻⁶ m²

Stress = 20/(3.142 x 10⁻⁶)

Stress = 6.37 x 10⁶ N/m²

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Erbium-165 has a half-life of 10.4 hours. If you start with 1,000 grams of
erbium-165, how much time will it take to have 125 grams of erbium-165 left
in the sample?
A. 41.6 hours
OB. 20.8 hours
C. 31.2 hours
D. 10.4 hours

Answers

The amount of time it will take to have 125 grams of erbium-165 left

in the sample is; C: 31.2 hours

How to calculate decay time?

The formula for amount of substance after decay is;

N(t) = N₀(¹/₂)^(t/t_¹/₂)

We are given;

Half life; t_¹/₂ = 10.4 hours

Initial amount; N₀ = 1000 g

Amount left; N(t) = 125 g

Thus;

125 = 1000 * (¹/₂)^(t/10.4)

125/1000 = (¹/₂)^(t/10.4)

In 0.125 = (t/10.4) In 0.5

-2.08 = -0.693(t/10.4)

t/10.4 = -2.08/-0.693

t = 10.4 * 3

t = 31.2 hours

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A force of looN acts upwards on a mass of 50 kg, another of
80N acts on the Same Mass downwards. Calculate the acceleration?
of the body and in which direction will be th acceleration?

Answers

Answer:

a force of acts on a body of mass 50 kg for 10 sec when the first of acting on the body the body covers 80 and in the next 10 second

Answer:

a = 0.4 ms⁻²

Explanation:

• First calculate the resultant force on the body:

If we consider upwards to be the positive direction and downwards to be the negative direction, then the resultant force will be:


F = +100 + (-80)

  = +20 N

The resultant force is positive, so it will act upwards.

• Next calculate the acceleration:
F = ma

a = F/m

a = 20/ 50

a = 0.4 ms⁻²

As the force is acting upwards, the acceleration will also be upwards.

According to the __________perspective, people who suffer from PTSD are classically conditioned.

Answers

According to the pavlovian perspective, people who suffer from PTSD are classically conditioned.

What is PSTD?

This is referred to as post traumatic stress disorder which is a mental disorder from a terrifying incident.

The pavlovian perspective however view this disorder as being a type of classical conditioned one.

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