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3 years ago

The speed at which an object falls and the acceleration at which it falls are

Physics

1 answer:

allochka39001 [22]3 years ago

6 0

No, the speed at which an object falls is not equal to the acceleration at which it falls.

Answer:

Option B

Explanation:

Speed is defined as how fast an object can cover a specific distance and in what time it covers. So it is measured as the ratio of distance covered to the time taken to cover that distance. While acceleration is the rate of change of velocity. Moreover, speed is a scalar quantity and acceleration is a vector quantity. So most of the times, the direction will play an important role in the varying values of speed and acceleration. Also, acceleration of an object will depend upon the force and mass of the object. Thus, speed and acceleration will not attain same value always.

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The left end of a long glass rod 8.00 cm in diameter and with an index of refraction of 1.60 is ground and polished to a convex

lesya692 [45]

Answer:

A) 0.1477

B) 0.65388 mm

C) object is inverted

Explanation:

The formula for object - image relationships for spherical reflecting surface is given as;

n1/s + n2/s' = = (n2 - n1)/R

Where;

n1 & n2 are the Refractive index of both surfaces

s is the object distance from the vertex of the spherical surface

s' is the image distance from the vertex of the spherical surface

R is the radius of the spherical surface

We are given;

index of refraction of glass; n2 = 1.60

s = 24 cm = 0.24 m

R = 4 cm = 0.04 m

index of refraction of air has a standard value of 1. Thus; n1 = 1

a) So, making s' the subject from the initial equation, we have;

s' = n2/[((n2 - n1)/R) - n1/s]

Plugging in the relevant values, we have;

s' = 1.6/[((1.6 - 1)/0.04) - 1/0.24]

s' = 0.1477

b) The formula for lateral magnification of spherical reflecting surfaces is;

m = -(n1 × s')/(n2 × s) = y'/y

Where;

m is the magnification

n1, n2, s & s' remain as earlier explained

y is the height of the object

y' is the height of the image

Making y' the subject, we have;

y' = -(n1 × s' × y)/(n2 × s)

We are given y = 1.7 mm = 0.0017 m and all the other terms remain as before.

Thus;

y' = -(1 × 0.1477 × 0.0017)/(1.6 × 0.24)

y' = - 0.00065388021 m = -0.65388 mm

C) since y' is negative and y is positive therefore, m = y'/y would result in a negative value.

Now, in object - image relationships for spherical reflecting surface, when magnification is positive, it means the object is erect and when magnification is negative, it means the object is inverted.

Thus, the object is inverted since m is negative.

6 0

3 years ago

You serve a volleyball with a mass of 0.77 kg. The ball leaves your hand with a velocity of

dimaraw [331]

<h2>Given :</h2>

mass (m) = 0.77 kg

velocity (v) = 2.33 m/s

<h2>Solution :</h2>

$\boxed{ \mathrm{Kinetic \: Energy = \frac{1}{2} m {v}^{2} }}$

$\dfrac{1}{2} \times 0.77 \times (2.33) {}^{2}$

$\dfrac{418.0253}{2}$

$209.01265 \: \: joules$

$\mathrm{209.01 \: \: joules} \: \: \: (approx)$

_____________________________

$\mathrm{ \#TeeNForeveR}$

7 0

3 years ago

n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend

Kitty [74]

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

$p_f=p_i\\mv_1+Mv_2=(m+M)v$

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

$(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}$

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

$E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2$

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

$(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m$

hence, the heigth is 68cm

4 0

3 years ago

Which statement accurately describes current electricity?

nikitadnepr [17]

Answer:

C???*

Explanation:

I forgot about this

4 0

3 years ago

Read 2 more answers

If a 400 watt motor is left on for 10 minutes, assuming no energy is lost to friction, how fast could a 100 kg cart be moving?

Verdich [7]

Answer:

69.28 m/s

Explanation:

From the question given above, the following data were obtained:

Power = 400 Watt

Time (t) = 10 minutes

Mass (m) = 100 Kg

Velocity (v) =?

Next, we shall convert 10 mins to seconds (s). This can be obtained as follow:

1 min = 60 s

Therefore,

10 mins = 10 × 60

10 mins = 600 s

Next, we shall determine the energy. This can be obtained as follow:

Power = 400 Watt

Time (t) = 600 s

Energy (E) =?

E = Pt

E = 400 × 600

E = 240000 J

Finally, we shall determine how fast the cart is moving. This can be obtained as illustrated below:

Mass (m) = 100 Kg

Energy (E) = Kinetic energy (KE) = 240000 J

Velocity (v) =?

KE = ½mv²

240000 = ½ × 100 × v²

240000 = 50 × v²

Divide both side by 50

v² = 240000 / 50

v² = 4800

Take the square root of both side

v = √4800

v = 69.28 m/s

Thus, the cart is moving with a speed of 69.28 m/s

5 0

3 years ago