Places rich in limestone and gypsum have hard water . why ??
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(a) The time spent in the air by the ball is 1.5 seconds.

(b) The horizontal distance of the ball is 9.83 m.

(c) The impact speed of the ball is 20.37 m/s.

(2) The magnitude of the force is 253.5 N in horizontal direction.

The time of motion of the ball is calculated as follows;

h = vt + ¹/₂gt²

18 = 8sin(35)t + (0.5)(9.8)t²

18 = 4.589t + 4.9t²

4.9t² + 4.589t - 18 = 0

solve the quadratic equation using formula method,

t = 1.5 s

X = vcosθ(t)

X = 8cos(35) x 1.5

X = 9.83 m

vyf = vyi + gt

vyf = 8sin(35) + 9.8(1.5)

vyf = 19.289 m/s

vxf = vxi

vxf = 8 x cos(35)

vxf = 6.55 m/s

Resultant speed = √(19.289² + 6.55²) = 20.37 m/s

F = ma

F = 65 x 3.9

F = 253.5 N

Since the motion is horizontal, the angle of the motion is zero.

Fx = 253.5cos(0) = 253.5 N

Fy = 253.5sin(0) = 0 N

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Answer:

D. 3 cm

Wavelength is the distance between two crests or two troughs.

In the given picture, the distance between two adjacent crests is 3 cm.

Hence, the wavelength is 3 cm.

Hence, the time required is approximately equal to 4 minutes and 39 seconds.

Given, Power, P = 300 W= 300J/s

Power= Energy per second

Mass, m= 0.25 kg

Initial Temperature= 20.0°C

Final Temperature= 100.0°C

Temperature difference, T =( 100-20)°C= 80.0°C

Specific heat of water, S= 4184 J/kg°C

Energy can be represented as

E=mST

where E is the energy, m is the mass, S is the specific heat of water and T is the temperature difference.

As Energy can be written as the product of time and power, E=Pt

Pt=mST

So,

t=(msT)/P

t=(0.25×4184×80)/300

Time, t = 278.934 seconds= 4 minutes and 39 seconds.

Hence, 4 minutes and 39 seconds will be required for a 300.0 W immersion heater to heat 0.25 kg of water from 20.0°C to 100.0°C.

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Answer:

Define universal gravitational constant. Why is newton's law of gravitation called universal law?

NGC 300 is different from our solar system because of its smaller size.

NGC 300 galaxy is different from our solar system because of its size. NGC 300 is smaller than our solar system which means that it has less number of stars as compared to our solar system.

So we can conclude that NGC 300 is different from our solar system because of its smaller size.

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The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.

The acceleration of the car before stopping at the given distance is calculated as follows;

v² = u² + 2as

when the car stops, v = 0

0 = u² + 2as

0 = 15² + 2(76.5)a

0 = 225 + 153a

-a = 225/153

a = - 1.47 m/s²

If the same force is applied, then acceleration is constant.

v² = u² + 2as

0 = 32² + 2(-1.47)s

2.94s = 1024

s = 348.3 m

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The impulse given to the ball by the floor is 0.2865 kg.m/s.

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.

A ball of mass 500g is dropped from a height 1.5m . It rebounds the floor and reached the height 1.2m.

The initial velocity u = √2x 9.81 x 1.5 = 5.425 m/s

The final velocity v = √2x 9.81 x 1.2 = 4.852 m/s

Substitute the values into the expression, we get

Impulse = m(v- u)

Impulse=0.5 x (4.852- 5.425 )

Impulse = - 0.2865 kg.m/s

Thus, the magnitude of impulse given to the ball by the floor is 0.2865 kg.m/s.

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Answer:equal mass

Explanation:the mass of both is 5kg

The mass in one period travels 'a distance less than 25 cm which will be 0.

The amplitude of a pendulum is the maximum displacement of the pendulum.

If the maximum displacement of the pendulum is 25 cm, the mass in one period travels a distance less that 25 cm.

Thus, then the mass in one period travels 'a distance less than 25 cm which will be 0.

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Two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.

When the body cell passes through cell division process, it divides into two daughter cells. In the second cell division, two cells divides into four cells. In the third cell division, the four cells divides into eight cells. In the fourth cell division, the eight cells divides and sixteen cells are produced.

So we can conclude that two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.

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Answer:

the answer to the question is a reflector

C. The Densities are equal.

Density is mass per unit volume or mass of a unit volume of a material substance.

If m1, V1 and D1 = mass, volume  and density respectively of ball C

m2, V2 and D2 = mass, volume and density respectively of ball D

According to the Question ,

[tex]V_{1} = 3V_{2} , m_{2} = \frac{1}{3} (m_{1} ) \\ \\= m_{1} = 3m_{2}[/tex]

Therefore,

[tex]\frac{D_{1} }{D_{2} } = (\frac{m_{1} }{V_{1} } )* (\frac{m_{2} }{V_{2} } )\\ \\= (\frac{3m_{2} }{3V_{2} })*(\frac{V_{2} }{m_{2} }) \\\\= 1[/tex]

Hence, D1 = D2

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Explanation:

this is the answer 276 volts

there is no pic or whatever to show

Answer:

D

Explanation:

GPE = mgh

GPE / gh = m

m = GPE/gh

m = 915 / 6.5 x 9.8

m = 915 / 63.7

m = 14.4kg

Answer:

54.0 x0.1=5.4 x0.03=0.162

kinetic force

If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03, then the minimum force F he must exert to get the block moving would be 5.2974 Nwrons.

Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.

As given in the problem, If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03,

The force required to get the block moving = μMg

                                                                          = 0.01×54×9.81

                                                                         = 5.2974 Newtons

Thus, the minimum force required to move the block would be 5.2974 Newtons.

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The true statement about the accuracy and precision of measuring tools is Tools cannot be both accurate and precise. The correct option is C.

When in an experiment, a value is measured 5 times, then if the values measured are same for most of the time or like three times out of five, it said to be accurate. The phenomenon is accuracy.

Precision is about comparing the values to each other then find them near to each other.

Accuracy compares the experimental value to the theoretical value.

So, when all the values are close to each other but not nearest to the theoretical value, then it is said to be precise but not accurate.

Thus, the correct option is C.

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Answer:

Since the momentum of the body remains constant ( conserved) the trolley slows down (its velocity reduces) since its mass increases.

The minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed without sleeping is 1.97.

The minimum coefficient of friction is calculated as follows;

F = μmg = mac

μmg = mac

where;

μg = ac

μg = v²/r

μ = v²/rg

where;

μ = (1.39²)/(0.1 x 9.8)

μ = 1.97

Thus, the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed without sleeping is 1.97.

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The charge on each sphere is 742 n

In physics, Charge, also known as electric charge is a electrical charge, or electrostatic charge and symbolized q, is a characteristic of a unit of matter that expresses the extent to which it has more or fewer electrons than protons.

According to the question,

Distance = 15cm or 0.15m

Force =  0.22N

[tex]K=9*10^{9} kgm^2/c^2[/tex] ( this is a constant value of K)

By using the formula,

F = [tex]\frac{kQq}{d^{2} }[/tex]

0.220 N = 8.99

N·m²/C² * Q² / (0.15m)²

Q = 7.42e-7

C = 742 n

C is the charge of each sphere .

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Let's see

[tex]\\ \rm\Rrightarrow mgh=\dfrac{1}{2}mv^2[/tex]

[tex]\\ \rm\Rrightarrow 2gh=v^2[/tex]

[tex]\\ \rm\Rrightarrow 2(10)(247)=v^2[/tex]

[tex]\\ \rm\Rrightarrow v²=4940[/tex]

[tex]\\ \rm\Rrightarrow v=70.3ms^{-1}[/tex]

The magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.

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 canonball= 47.4 m/sec

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

v is the velocity after 1.23 s

According to Newton's first equation of motion,

[tex]\rm v=u +gt \\\\ v= 47.4 \ m/sec + 9.81 \ m/sec^2 \times 1.23 sec \\\\\ v=59.46 \ m/sec[/tex]

Hence, the magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.

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Gravity and normal forces act on the toy car.

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Answer:

suna bro i am nepali and nepali are pro

The height of the object will be -5.19 cm

A concave mirror's reflecting surface curves inward and away from the light source. Light is reflected inward to a single focus point via concave mirrors. Concave mirrors, in contrast to convex mirrors, produce a variety of images depending on the object's to the mirror.

Given an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall

So let,

v =  Image distance from the mirror = -33.5 cm

u = object distance from the mirror (concave) = 24 cm

hi = Image height = 7.25 cm

h = height of the object = ?

Using below formula to find height of the object

-v/u = hi/h

Putting all value in the formula we get

-(-33.5)/(-24) = 7.25/h

h = -5.19 cm

Therefore the height of the object will be -5.19 cm

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The time for which the baseball remains in the air will be 9.37 sec. Option C is correct.

It is the height achieved by the body when a body is thrown at the same angle and the body is attaining the projectile motion.

From Newton's first equation of motion;

[tex]\rm v_y= u_y+gt \\\\ 0 = usin \theta - gt \\\\ t = \frac{u sin\theta }{g}\\\\t = 4.68 \ sec[/tex]

The time the baseball remains in the air will be twice the time he travels for one side motion;

T' = 2t

T'=2 × 4.68 sec

T'= 9.37 sec

Hence for 9.37 sec, the baseball willl remains in the air. Option C is correct.

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We can obtain the velocity from the graph and plot is against time to obtain a velocity time graph.

The velocity time graph shows the change of velocity with time. Recall that velocity is the change of position with time.

As such, when we have a position time graph as shown, we can obtain the velocity from the graph and plot is against time to obtain a velocity time graph.

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He would measure the speed of molecules as the temperature gets colder until the temperature is measured at which motion stops(option A).

The absolute zero refers to the temperature at which the gas seems to stop moving. Recall that temperature is the measure of the average kinetic energy of the molecules of a gas. We now have to consider the options and see the one that connects temperature and speed of molecules.

Hence, if Guillaume Amontons first took a stab at measuring absolute zero in 1702, he would measure the speed of molecules as the temperature gets colder until the temperature is measured at which motion stops.

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The total time required is 1 minute and 0.3 seconds.

Assumption: The distance between the people is 1 m and the stadium is a circle with a radius of 100m.

Here, the time taken by the person is 0.1 seconds.

Total distance covered by the wave = Circumference of the circle

So, total distance = 2 πr = 2 × 3.14 × 100 = 618 m

As the distance between each of the people is 1 m.

So, the number of personal interactions is 618.

Time taken by each person is 0.1 seconds.

So, total time, t = 0.1 × 618 = 61.8 seconds.

So, the order of the magnitude of the time required is 1 minute and 0.3 seconds for such a pulse to make one circuit around BC Place Stadium in Vancouver.

Hence, the total time required is 1 minute and 0.3 seconds.

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(a) The acceleration of the two block is determined as 0.93 m/s².

(b) The tension in the string is 5.78 N.

The net force on the aluminum block is calculated as follows;

[tex]T - \mu_a m_a g = m_a a \ --- \ (1)[/tex]

The net force on the copper block is calculated as follows;

[tex]m_cg sin(30) - T -\mu _cm_c gcos(30) = m_c a --- (2)[/tex]

where;

Solve for T using (1) and (2)

[tex]a = g(\frac{m_c sin30\ - \ \mu _c m_c cos30 \ - \mu_ a m_a}{ma_a + m_c} )\\\\a = 9.8(\frac{6 sin30\ - \ 0.2 (6) cos30 \ - 0.2 (6)}{2 +6} )\\\\a = 0.93 \ m/s^2[/tex]

From equation (1);

T = μm_ag + m_aa

T = 0.2(2)(9.8) + 2(0.93)

T = 5.78 N

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