1. Two forces F~
1 and F~
2 are acting on a block of mass m=1.5 kg. The magnitude of
force F~
1 is 12N and it makes an angle of θ = 37◦ with the horizontal as shown in
figure-1. The block is sliding at a constant velocity over a frictionless floor.
Figure-1
(a) Find the value of the normal force on the block.
[3 marks]
(b) Find the magnitude of force F~
2 that is acting on the block
[4 marks]
(c) Find the magnitude of force F~
2 if the block accelerates with a magnitude of
a = 2.5 m/s2 along the direction of F~
2 . [3 mark

Answer:

big bang theory

Explanation:

Edwin Hubble is credited for the initial development of the Big Bang theory, an idea which helps to explain the formation of the universe over 15 billion years ago.

Answer:

ok

Explanation:

ok ok I will thanks for free

Answer:

0.01 s

Explanation:

When the capacitor voltage is maximum, the inductor voltage is at a minimum, and when the capacitor voltage is minimum, the inductor voltage is at a maximum.

The time difference between minimum voltage and maximum voltage for the capacitor or inductor is the same as that for maximum voltage of capacitor to maximum voltage of inductor and this is t = T/4 where T = period.

Since T = 0.04 s, t = T/4

= 0.04 s/4

= 0.01 s

So, the amount of time is there between when the capacitor voltage maximizes and the inductor voltage maximizes is 0.01 s

Answer:

During a collision, some of the ball's energy is converted into heat. As no energy is added to the ball, the ball bounces back with less kinetic energy and cannot reach quite the same height. The heavy ball, on the other hand, is left behind with little energy and does not move much.

Answer:

body mass index

Explanation:

Answer:

[tex] \boxed{ \sf{ \: Option \: C \: is \: correct.}}[/tex]

Explanation:

BMI stands for Body Mass Index .

Hope I helped!

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~TheAnimeGirl

Answer:

you will die by it

Explanation:

Answer:

If the intake stroke never occurred, then the whole four stoke enegine process would not work

Explanation:

Answer:

I believe none

Explanation:

Answer:

The minimum downward force on the box that will keep it from slipping is 63.14 N

Explanation:

Given;

mass of the object, m = 30 kg

applied force, f = 125 N

coefficient of static friction, μ = 0.35

Normal reaction (R) is acting upwards, weight of the box (mg) is acting downwards and the minimum downward force (F) on the box that will keep it from slipping is also acting downwards.

The net vertical forces on the box is given by;

R - mg - F = 0

F = R - mg

Now, determine normal reaction, R

f = μR

R = f / μ

R = 125 / 0.35

R = 357.14 N

Finally, determine the minimum downward force on the box that will keep it from slipping;

F = R - mg

F = 357.14 - (30 x 9.8)

F = 357.14 - 294

F = 63.14 N

Therefore, the minimum downward force on the box that will keep it from slipping is 63.14 N

Answer:

[tex]\mid \vec C\mid=31.9[/tex]

Explanation:

Consider the vectors:

[tex]\vec A=9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}}[/tex]

[tex]\vec B=-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}}[/tex]

Calculate the magnitude of

[tex]\vec C=-2\vec B-\vec A[/tex]

Substitute the values of the vectors:

[tex]\vec C=-2(-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}})-(9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}})[/tex]

Operate and remove parentheses:

[tex]\vec C=19\mathbf{\hat{i}}+26.8\mathbf{\hat{j}}-9.4\mathbf{\hat{i}}+3.6\mathbf {\hat{j}}[/tex]

Operating both components separately:

[tex]\vec C=9.6\mathbf{\hat{i}}+30.4\mathbf{\hat{j}}[/tex]

Now find the magnitude of C:

[tex]\mid \vec C\mid=\sqrt{9.6^2+30.4^2}[/tex]

[tex]\mid \vec C\mid=\sqrt{1016.32}[/tex]

[tex]\mathbf{\mid \vec C\mid=31.9}[/tex]

Answer:

you need to multiply the momentum and the mass

Answer:

2.5 seconds

Explanation:

Time is equal to distance over speed

300 divide by 120 u get 2.5 seconds

Answer:

[tex]\mathbf{F_{thrust} \simeq 2.3 \times 10^4 \ N}[/tex]

Explanation:

For a lunar lander acceleration in an upward direction, the net force that acts on it can be expressed as:

[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex]

The equation for the net force that acts on the lunar lander is:

[tex]F_{net} = m_La[/tex]

For the lander touch down to be zero acceleration, it implies that the acceleration of the moving rocket is zero(a free body fall)

i.e. a = 0

We can now regard the Apollo 12 lunar as a freely falling body

However; the force of gravity as a result of the moon acting on the lunar rocket is:

[tex]F_{g(moon)} = m_Lg_m[/tex]

Then; the equation for the thrust force of the lunar rocket is:

[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex]

[tex]m_La = F_{thrust}-m_Lg_m[/tex]

[tex]m_L(0)= F_{thrust}-m_Lg_m[/tex]

[tex]0= F_{thrust}-m_Lg_m[/tex]

[tex]-F_{thrust}= -m_Lg_m[/tex]

[tex]F_{thrust}= m_Lg_m[/tex]

Finally; the thrust force of the lunar rocket can be determined as:

[tex]F_{thrust}= m_Lg_m[/tex]

Acceleration due to gravity ot the surface of the moon = 1.625 m/s²

[tex]F_{thrust}=(1.4 \times 10^{4} \ kg) (1.625 \ m/s^2)[/tex]

[tex]F_{thrust}=2.275 \times 10^4 \ N[/tex]

[tex]\mathbf{F_{thrust} \simeq 2.3 \times 10^4 \ N}[/tex]

A thrust force of 22680N will be necessary to  to have the lander touch down with zero acceleration.

Given the data in the question;

Mass of Apollo 12 lunar lander; [tex]m_L = 1.4 * 10^4 kg[/tex]

Using the expression for the net-force that acts on the lander when it is accelerated in the upward direction:

[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex] ------- Let this be equation 1

Also, the net-force that acts on the lunar lander can be expressed as:

[tex]F_{net} = m_L*a[/tex] ------- Let this be equation 2

Where [tex]m_L[/tex] is the mass of the lunar lander and a is the acceleration ( 0 ),

( the acceleration of the moving rocket is zero and its freely falling )

Also, Force of gravity due to moon that acts on the lunar lander is expressed as:

[tex]F_{g(moon)} = m_L * g_m[/tex]  ------- Let this be equation 3

Where [tex]m_L[/tex] is the mass of the lunar lander and [tex]g_m[/tex] is the moon's gravity ( [tex]1.62m/s^2[/tex] )

Lets substitute equation 2 and 3 into 1

[tex]m_L * a = F_{thrust} - (m_L * g_m)\\\\m_L * 0= F_{thrust} - (m_L * g_m)\\\\0=F_{thrust} - (m_L * g_m)\\\\ F_{thrust} = m_L * g_m[/tex]

We substitute our values into the equation

[tex]F_{thrust} = (1.4*10^4kg) * 1.62m/s^2\\\\F_{thrust} = 22680 kg.m/s^2\\\\F_{thrust} = 22680N[/tex]

Therefore, a thrust force of 22680N will be necessary to  to have the lander touch down with zero acceleration.

Learn more: https://brainly.com/question/7602665

Answer:

real world examples

Explanation:

concave wave rock- The wave rock is formed by weathering of the surrounding area, which helps in proving the deposition part, as the wave rock was below the ground, occurred due to deposition of rock years over years. It is made from very tough material from its surrounding. The weathering reduced the surrounding terrain, while the bedrock remained to witness for the history.  --Also, volcanic eruptions have changed earth greatly. EX- Ash and sulfur went into Earth's atmosphere because of the Tambora eruption which dimmed incoming sunlight. It lowered global temperatures by about 3°F. The Mount Pinatubo of 1991 that erupted in the Philippines cooled the planet by about 1°F.

Answer:

I'm going to guess 15 N

Explanation:

Answer:

Magnitude of gravitational force between this planet and its moon: approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].

Acceleration of this moon towards the planet: approximately [tex]5.49 \times 10^{-3}\; \rm m \cdot s^{-2}[/tex].

Acceleration of this planet towards its moon: approximately [tex]2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].

Explanation:

Look up the gravitational constant, [tex]G[/tex]:

[tex]G \approx 6.67 \times 10^{-11}\; \rm m^3\cdot kg^{-1} \cdot s^{-2}[/tex].

Assume that both this planet and its moon are spheres of uniform density. When studying the gravitational interaction between this planet and its moon, this assumption allows them to be considered as two point masses.

The formula for the size of gravitational force between two point masses [tex]m_1[/tex] and [tex]m_2[/tex] with a distance of [tex]r[/tex] in between is:

[tex]\displaystyle F = \frac{G \cdot m_1 \cdot m_2}{r^2}[/tex],

where [tex]G[/tex] is the gravitational constant.

Let [tex]m_1[/tex] and [tex]m_2[/tex] denote the mass of this planet and its moon, respectively.

Calculate the size of gravitational force between this planet and its moon:

[tex]\begin{aligned} F &= \frac{G \cdot m_1 \cdot m_2}{r^2} \\ &\approx \frac{6.67 \times 10^{-11}\; \rm m^3 \cdot kg^{-1} \cdot s^{-2} \times 6.00 \times 10^{24}\; \rm kg \times 2.50 \times 10^{22}\; \rm kg}{{\left(2.70 \times 10^{8}\; \rm m\right)}^2} \\ &\approx 1.37 \times 10^{20}\; \rm N\end{aligned}[/tex].

Assume that other than the gravitational force between this planet and its moon, all other forces (e.g., gravitational force between this planet and the star) are negligible. The magnitude of the net force on the planet and on the moon should both be approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].

Apply Newton's Second Law of motion to find the acceleration of this planet and its moon:

[tex]\displaystyle \text{acceleration} = \frac{\text{net force}}{\text{mass}}[/tex].

For this moon:

[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{2.50 \times 10^{22}\; \rm kg} \approx 5.49\times 10^{-3}\; \rm m \cdot s^{-2}[/tex].

For this planet:

[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{6.00 \times 10^{24}\; \rm kg} \approx 2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].

Answer:

1 hour

Explanation:

t = 6/6

6/6 = 1

You just said that his speed enables him to cover 6 miles in each hour. From this information, I calculated that it will take him one hour to travel 6 miles.

We are given:

The tuning fork vibrates at 15660 oscillations per minute

Period of one back-and forth movement:

the given data can be rewritten as:

1 minute / 15660 oscillations

60 seconds / 15660 oscillations          (1 minute  = 60 seconds)

dividing the values

0.0038 seconds / Oscillation

Therefore, one back and forth vibration takes 0.0038 seconds

Answer:

d = 7.5 [cm]

Explanation:

To be able to solve this problem we must understand well the concept of a simple lever of First Class, which consists of a support point (fulcrum) and a beam where a force is applied at one end, while at the other end the force exerted is multiplied.

In the attached image we can see an image that will help us better understand the concept of this first-class lever.

The moment created by the effort and the distance is equal to:

M = 100*30 = 3000 [N*cm]

Now the balance is created because we have the same moment exerted by the effort and the one created by the load.

3000 = 400*d

d = 7.5 [cm]

Answer:

The value is [tex]B = 0.2312 \ T[/tex]

The direction is into the surface

Explanation:

From the question we are told that

   The mass density is  [tex]\mu =\frac{m}{L} = 1.17 \ g/cm =0.117 kg/m[/tex]

   The coefficient of kinetic friction is  [tex]\mu_k = 0.250[/tex]

   The current the wire carries is  [tex]I = 1.24 \ A[/tex]

Generally the magnetic force acting on the wire is mathematically represented as

         [tex]F_F = F_B[/tex]

Here   [tex]F_F[/tex] is the frictional  force which is mathematically represented as

      [tex]F_F = \mu_k * m * g[/tex]

While [tex]F_B[/tex]  is the magnetic force which is mathematically represented as

       [tex]F_B = BILsin(\theta )[/tex]

Here [tex]\theta =90^o[/tex] is the angle between the direction of the force and that of the current

So

      [tex]F_B = BIL[/tex]

So

      [tex]BIL = \mu_k * m * g[/tex]

=>   [tex]B = \mu_k * \frac{m}{L} * [\frac{g}{I} ][/tex]

=>   [tex]B = 0.25 * 0.117 * [\frac{9.8}{1.24} ][/tex]

=>   [tex]B = 0.2312 \ T[/tex]

Apply the right hand curling rule , the thumb pointing towards that direction of the current we see that the direction of the magnetic field is into the surface as shown on the first uploaded image

Answer:

D

Explanation:

Answer:

If you throw a pebble into a pond, ripples

spread out from where it went in. These

ripples are waves travelling through the

water. The waves move with a transverse

motion.

Explanation:

Answer:

The ball will hit the ground 108.25 m down the field

Explanation:

To determine how far down the field the ball will hit the ground, that is the Range of the ball. From formula to calculate Range in projectile motion,

[tex]R = \frac{u^{2} sin2\theta }{g}[/tex]

Where R is the Range

u is the initial velocity

θ is the angle of projection

and g is acceleration due to gravity (Take g = 9.8 m/s²)

From the question,

Initial velocity, u = 35 m/s

Angle, θ = 60°

Putting these values into the equation, we get

[tex]R = \frac{35^{2} ( sin2(60)) }{9.8}[/tex]

[tex]R = \frac{1225 \times sin(120)}{9.8}[/tex]

[tex]R = \frac{1225 \times 0.8660}{9.8}[/tex]

[tex]R = \frac{1060.85}{9.8}[/tex]

[tex]R = 108.25 m[/tex]

Hence, the ball will hit the ground 108.25 m down the field.

Answer:

Explanation:

The total work done by the wave is expressed as;

Workdone = Potential energy + Kinetic energy

Workdone = mgh + 1/2mv²

m is the mass = 77kg

g is the acceleration due to gravity = 9.8m/s²

v is the velocity = 8.2m/s

h is the height = 1.65m

Substitute into the formula;

Workdone = 77(9.8)(1.65) + 1/2(77)8.2²

Workdone = 1245.09 + 2588.74

Workdone = 3833.83Joules

Hence the amount of non conservative work done on the sofa is 3833.83Joules

Given:

We know,

→ [tex]Work \ done = Potential \ energy +Kinetic \ energy[/tex]

or,

                     [tex]= mgh +\frac{1}{2} mv^2[/tex]

By putting the values,

                     [tex]= 77\times 9.8\times 1.65+\frac{1}{2}\times 77\times (8.2)^2[/tex]

                     [tex]= 1245.09+2588.74[/tex]

                     [tex]= 3833.83 \ Joules[/tex]

Thus the above approach is right.

Learn more about work done here:

https://brainly.com/question/24230840

At angular speed of 3 rev/s, the object moves a distance equal to 3 times the circumference of the circle each second, or a distance of 3 • 2π (1.20 cm) ≈ 22.6 cm.

So, with a linear speed of 22.6 cm/s = 0.226 m/s, the object has a centripetal acceleration a of

a = (0.226 m/s)² / (0.012 m) ≈ 18.8 m/s²

directed toward the center of the circle.

Answer:

J

Explanation:

The daughter moves with greater acceleration backwards because of her weight.

Both mother and daughter move backward, but the daughter moves with greater acceleration. Therefore, option (J) is correct.

This law states that the acceleration of a body depends on two factors. The first one is the mass of the body and the net force acting on the body. The acceleration is inversely proportional to the mass and is directly proportional to the net force acting on the body and

Newton’s second law can be written in an expression as follows:

F = ma

or,   a = F/m

According to Newton's third law of motion, there is an equal and opposite reaction for every action. So when the two skaters push each other. They both will move in a backward direction.

But the mass of the daughter is 50 Kg while the mass of the mother is 100 Kg. As the mass of the daughter is less. Therefore daughter moves in a backward direction with greater acceleration.

Therefore, both move backward but the daughter moves with greater acceleration in comparison to the mother.

Learn more about Newton's second law of motion, here:

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#SPJ2

Answer:

OK?

Explanation:

Answer:

It happens and they might not even cook what you feel like eating.

Answer:

The average coefficient of static friction = 0.130

Explanation:

The average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]

From the given question, the coefficient of static friction given are:

0.053, 0.081, 0.118, 0.149, 0.180, 0.198

Thus,

the sum of the given coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                             = 0.779

The average coefficient of static friction = [tex]\frac{0.779}{6}[/tex]

                                             = 0.129833

                                             = 0.130

Therefore, the average coefficient of static friction is 0.130.