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vekshin1
3 years ago
15

The hubbles telescopes orbit is 5.6 x10 ^5 meters above earths suface. the telescope has a mass os 1.1 x10^4 kilograms. earth ex

erts a gravitational force of 9.1 x10^4 newtons on the telescope. what is the magnitude of earths gravitational field strength at this location
Physics
1 answer:
Andru [333]3 years ago
7 0
(3) 8.3 N/kg. The gravitational field strength at a point is the force per unit mass exerted on a mass placed at that point. So at the point where the Hubble telescope is, it is (9.1 x 10^4)N/(1.1 x 10^4 kg) = 8.3 N/kg

Fam
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Now explore friction force. Set the piece of plastic or wood on the table and push it steadily across the tabletop using your fi
ehidna [41]

Answer:

The tabletop is smooth so my finger is down it fast and easy. The fabric however slowed my finger down considerably, and it was harder for me to move my finger across it.

Explanation:

Hope this helps.

3 0
2 years ago
Why do some objects fall faster than others?
monitta

Answer:

Weight and Mass !!!!!!

Explanation:

Galileo discovered that objects that are more dense, or have more mass, fall at a faster rate than less dense objects, due to this air resistance. A feather and brick dropped together. Air resistance causes the feather to fall more slowly.

3 0
3 years ago
Plane a travels at 900km/h and plane b travels at 250/5.which plane travels faster
vesna_86 [32]

Explanation:

We have,

Speed of plane a is 900 km/h

Plane b is moving at a rate of \dfrac{250\ km}{5\ h}=50\ km/h

It is required to find which plane is faster. To find which plane is faster, we need to compare their speeds.

Speed of a plane a is 900 km/h and that of plane b is 50 km/h. So, we can say that plane a is moving faster.

5 0
3 years ago
A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur
morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\

\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] =  4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\

Substitute the given parameters and solve for the angular speed;

\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\  \ +\  \ 0.35(0.02^2\  + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0
2 years ago
A bike accelerates uniformly(from rest to a speed of 7 m/s over a distance of 40 m. Determine
marshall27 [118]

Answer:

0.61 m/s^2

Explanation:

The bike's acceleration can be found by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity of the bike

u is the initial velocity

a is the acceleration

s is the distance covered

For the bike in the problem,

u = 0

v = 7 m/s

d = 40 m

Solving the equation for a, we find the acceleration:

a=\frac{v^2-u^2}{2s}=\frac{7^2-0}{2(40)}=0.61 m/s^2

3 0
3 years ago
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