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marissa [1.9K]
3 years ago
10

You are interested in space travel and want to learn more facts about space travel. Which would be the best source of informatio

n? (2 points)
Physics
2 answers:
Tpy6a [65]3 years ago
5 0
A book or a professor
Citrus2011 [14]3 years ago
5 0

The correct answer would be NASA as it will be proven

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Juan rides his horse with a constant<br> speed of 18 km/h. How far can he travel<br> in 1/2 hour?
Brut [27]

Answer:

He traveled 9km

Explanation:

To do this problem you need to use the equation which is Speed= distance/time and this problem gives you the speed which is 18 km/h and it gives you the time 1/2 hour so you write the equation 18= d/ 1/2 which his distance is 9km

3 0
3 years ago
to what height will a 250g soccer ball rise to if it is kicked directly upwards at 8 meters per second​
Nonamiya [84]

Answer:

3.2 m

Explanation:

we know

Hmax = V²/2g

= 8² / 2*10 = 3.2

6 0
3 years ago
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How is work involved in stopping a car?
vichka [17]
You have to move your foot to stop the car so I guess that would be considered work by moving your foot
5 0
3 years ago
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You must exert a force of 4.5 N on a book to slide it across a table. If you do 2.7J
hoa [83]

Answer: 0.6m

Explanation:

Given that:

force = 4.5 N

Work done = 2.7J

Distance moved by the book = ?

Since work is done when force is applied on an object over a distance, apply the formula:

work = force x distance

2.7J = 4.5N x distance

Distance = (2.7J / 4.5N)

Distance = 0.6 m

Thus, the book was moved 0.6 metres far

3 0
3 years ago
An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic m
Elanso [62]

Answer:

Part a)

A = 0.066 m

Part b)

maximum speed = 0.58 m/s

Explanation:

As we know that angular frequency of spring block system is given as

\omega = \sqrt{\frac{k}{m}}

here we know

m = 3.5 kg

k = 270 N/m

now we have

\omega = \sqrt{\frac{270}{3.5}}

\omega = 8.78 rad/s

Part a)

Speed of SHM at distance x = 0.020 m from its equilibrium position is given as

v = \omega \sqrt{A^2 - x^2}

0.55 = 8.78 \sqrt{A^2 - 0.020^2}

A = 0.066 m

Part b)

Maximum speed of SHM at its mean position is given as

v_{max} = A\omega

v_{max} = 0.066(8.78) = 0.58 m/s

4 0
3 years ago
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