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viktelen [127]
2 years ago
6

A space station that does not rotate cannot simulate gravity for its occupants.

Physics
1 answer:
il63 [147K]2 years ago
4 0

false

true

false

false

true

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An object is at rest on the ground. The object experiences a downward gravitational force from Earth. Which of the following pre
Montano1993 [528]

Answer:

A) and B) are correct.

Explanation:

If the object is at rest, it means that no net force is exerted on it.

As the object experiences a downward gravitational force from Earth, in order to be at rest, it must experience an upward force with the same magnitude as the gravitational force on the object.

This force is supplied by the normal force, which can adopt any value in order to meet the condition imposed by Newton´s 2nd Law, and is always perpendicular to the surface on which the object is placed (in this case, the ground).

At a molecular level, this normal force is supplied by the bonded molecules of the ground that behave like small springs being compressed by the molecules of the object, exerting an upward restoring force upward on them.

So, the statements A) and B) are true.

6 0
3 years ago
Read 2 more answers
A 150 kg bike takes a roundabout with a radius of 53.0 m. The roundabout it’s only ¾ of a circle. the time taken was 0.35 minute
BigorU [14]

Answer:

ef<grhtxdgfkugılohşrgzfxthyjhuk

Explanation:dfbgdhzfgbfxgzbfxd

fadwgfrsdfkyugılhşojipdaftshrthsxxfbzxryfchtgghxfjnx hjnnvbm bn dfkunggvxbvbbfbfxxb

5 0
3 years ago
How fast can a 4000 kg truck travel around a 70 m radius turn without skidding if its tires share a 0.6 friction coefficient wit
aleksley [76]

Answer:

Velocity of truck will be 20.287 m /sec

Explanation:

We have given mass of the truck m = 4000 kg

Radius of the turn r = 70 m

Coefficient of friction \mu =0.6

Centripetal force is given  F=\frac{mv^2}{r}

And frictional force is equal to F_{frictional}=\mu mg

For body to be move these two forces must be equal

So \frac{mv^2}{r}=\mu mg

v=\sqrt{\mu rg}=\sqrt{0.6\times 70\times 9.8}=20.287m/sec

7 0
2 years ago
Calculate the magnitude of the linear momentum for the following cases. (a) a proton with mass 1.67 10-27 kg, moving with a spee
abruzzese [7]

Answer:

<h2>8.0995×10^-21 kgms^-1</h2>

Explanation:

Mass of proton :

m_P=1.67\times 10^-^2^7\:kg\\

Speed of Proton:

v_P=4.85\times 10^6

Linear Momentum of a particle having mass (m) and velocity (v) :

-> p =m->v\:\:\: (1)

Magnitude of momentum :

p=mv\:\:\: (2)

Frome equation (2), magnitude of linear momentum of the proton :

p_P=m_P\:v_P\\\\p_P=1.67\times 10^-^2^7 \:kg\times4.85\times 10^6\:ms^-^1\\\\p_P= 8.0995\times 10^-^2^1\:kgms^-^1

7 0
2 years ago
Amount of force exerted on an object due to gravity is called
MAXImum [283]
The gravitational force between a mass and the Earth is the object'sweight. Mass is considered a measure of an object's inertia, and its weight is the force exerted on the object in a gravitational field. On the surface of the Earth, the two forces are related by the acceleration due to gravity: Fg = mg.

Hoped this helped!
6 0
3 years ago
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