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bearhunter [10]

2 years ago

9

A woman drives a 45 gram golf ball off of a tee and gives the ball a velocity of 28 m/s to the right. What is the change in mome

ntum of the ball?

1 answer:

lidiya [134]2 years ago

7 0

Answer:

1.26

Explanation:

change 45 grams to kilograms which is 0.045 then just multiply it by 28 m/s

0.045x28=1.260

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An object of mass 0.40 kg, hanging from a spring with a spring constant of 8.0 N/m, is set into an up-and-down simple harmonic m

Sergeeva-Olga [200]

Answer:

a = 2 m/s2

Explanation:

we know from newtons 2nd law

F = ma.

we also know that from hookes law we have

F = kx

equate both value of force to get value of acceleration

kx = ma,

where,

k is spring constant = 8.0 N/m

x is maximum displacement 0.10 m

m is mass of object 0.40 kg

a = \frac{kx}{m}

= \frac{8 *0 .10}{0.40}

a = 2 m/s2

5 0

3 years ago

Astronauts aboard the ISS move at about 8000 m/s, relative to us when we look upward.How long does an astronaut need to stay abo

Luba_88 [7]

Answer:

#_time = 7.5 10⁴ s

Explanation:

In order for the astronaut to be younger than the people on earth, it follows that the speed of light has a constant speed in vacuum (c = 3 108 m / s), therefore with the expressions of special relativity we have.

t = $\frac{t_p}{ \sqrt{1- (v/c)^2} }$

where t_p is the person's own time in an immobile reference frame,

$t_{p} = t \sqrt{1 - (\frac{v}{c})^2 }$

let's calculate

we assume that the speed of the space station is constant

$t_p = 1 \sqrt{1 - \frac{8 \ 10^3}{3 \ 10^8} }$

$t_p = 1 \sqrt{1- 2.6666 \ 10^{-5}}$

t_ = 0.99998666657 s

therefore the time change is

Δt = t - t_p

Δt = 1 - 0.9998666657

Δt = 1.3333 10⁻⁵ s

this is the delay in each second, therefore we can use a direct rule of proportions. If Δt was delayed every second, how much second (#_time) is needed for a total delay of Δt = 1 s

#_time = 1 / Δt

#_time = $\frac{1}{1.3333 \ 10^{-5}}$

#_time = 7.5 10⁴ s

5 0

3 years ago

A mountain climber increases their height from 200 meters to 400 meters. What affect will this have on their potential energy?

Yanka [14]

Answer:

At 400 m the potential energy of the mountain climber doubled the initial value.

Explanation:

Given;

initial height of the mountain climber = 200 m

final height of the mountain climber, = 400 m

The potential energy of the mountain climber is calculated as;

Potential energy, P.E = mgh

At 200 m, P.E₁ = mg x 200 = 200mg

At 400 m, P.E₂ = mg x 400 = 400mg

Then, at 400 m, P.E₂ = 2 x 200mg = 2 x P.E₁

Therefore, at 400 m the potential energy of the mountain climber doubled the initial value.

4 0

3 years ago

Light of wavelength λ travels through a medium with an index of refraction n1before striking a thin film with an index of refrac

ra1l [238]

Answer:

option C

Explanation:

The correct answer is option C

A light that transmits through n₂ travels t distance before reflection off the n₁ medium and again travels distance t before reaching the point from where it entered n₂ medium. Hence it travels 2 t distance more than the light that is reflected off n₂.

It( light entering n₂) also travels an additional distance equal to, half of the wavelength, when reflected off n₁ ( as n₁ is greater than n₂).

Wavelength in n₂ is = $\dfrac{\lambda}{n_2}$

Hence, path length difference = $2t +\dfrac{\lambda}{2 n_2}$

5 0

3 years ago

A satellite of mass 5600 kg orbits the Earth and has a period of 6200 s.

Troyanec [42]

Answer:

(a) Radius of orbit will be $=7.32\times10^6m$

(b) Earth gravitational force will be $=4.18\times 10^4N$

(C) Height will be $0.92\times 10^6m$

Explanation:

We have given

Mass of the earth, $M=6\times 10^{24}kg$

Mass of the satellite, m = 5600 kg

Radius of earth, $R=6.4\times 10^6m$

Time period T = 6200 sec

We know that $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{6200}=0.00101rad/sec$

Now

(a) We know that $\omega ^2=\frac{GM}{R^3}$

$R^3=\frac{GM}{\omega ^2}$

$R^3=\frac{6.67\times 10^{-11}\times 6\times 10^{24}}{0.00101 ^2}$

$R^3=3.92\times 10^{20}$

Radius of the orbit $R=7.32\times 10^6m$

(b)

Force $F=\frac{GMm}{R^2}=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 5600}{(7.32\times 10^6)^2}=4.18\times 10^4N$

(c)

Altitude $h=radius\ of\ orbit-radius\ of\ earth=7.32\times 10^6-6.4\times 10^6=0.92\times 10^6m$

8 0

3 years ago