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3 years ago

Select all the correct answers.

Physics

1 answer:

Anastasy [175]3 years ago

8 0

The ground exerts an equal force on the golf ball

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4. A car accelerates, from rest, at 2.4 m/s?. How fast is the car traveling

pantera1 [17]

Answer:

19.2m/s

Explanation:

Assuming that 2.4m/s^2 was the acceleration and not a typo, we can use the equation v=at, where v=velocity, a=acceleration, and t=time,

plug in known varibles,

v=2.4*8

v=19.2m/s

4 0

3 years ago

What is the net force exerted on a 28.8 kg shopping cart that accelerates at a rate of 2.88 m/s^2?

Mama L [17]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 28.8 × 2.88 = 82.944

We have the final answer as

Hope this helps you

6 0

3 years ago

Sonja [21]

Answer

2) 1.5×10-2 m

Explanation

The potential difference is related to the electric field by:

$\Delta V=Ed$ (1)

where

$\Delta V$ is the potential difference

E is the electric field

d is the distance

We want to know the distance the detectors have to be placed in order to achieve an electric field of

$E=1 V/cm=100 V/m$

when connected to a battery with potential difference

$\Delta V=1.5 V$

Solving the equation (1) for d, we find

$d=\frac{\Delta V}{E}=\frac{1.5 V}{100 V/m}=0.015 m=1.5 \cdot 10^{-2} m$

5 0

3 years ago

If the mass of a 1.8 g paperclip was able to be completely converted to energy, how much energy would you obtain?

Anton [14]

Answer:

$E=1.62\times 10^{14}\ J$

Explanation:

Given that,

The mass of the paperclip, m = 1.8 g = 0.0018 kg

We need to find the energy obtained. The relation between mass and energy is given by :

$E=mc^2$

Where

c is the speed of light

So,

$E=0.0018\times (3\times 10^8)^2\\\\E=1.62\times 10^{14}\ J$

So, the energy obtained is $1.62\times 10^{14}\ J$ .

7 0

3 years ago

A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal

slamgirl [31]

Answer:

18.89cm

Explanation:

As we know that the person is standing 5m in front of the camera

$d_0=5m=500cm$

The focal length of the lens =50cm

f=50 cm

By Lens formula we have:

$\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm$

By the formula of magnification

$\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm$

The height of the image formed is 18.89cm.

5 0

3 years ago