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vodomira [7]
4 years ago
9

Acceleration can bea change in speed and.........?

Physics
1 answer:
statuscvo [17]4 years ago
4 0

Acceleration is any change in speed or direction of motion.

Speeding up, slowing down, or moving along a curve are all accelerations.


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Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat
svetlana [45]

Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

g = GM/r²

G is the gravitational constant

M is the mass = 1.0 x 10^5 kg

r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

6 0
2 years ago
alculate the kinetic energies of (a) a 2.00×103-kg automobile moving at 100.0 km/h; (b) an 80.0-kg runner sprinting at 10.0 m/s;
zzz [600]

Answer:

(a) 7.72×10⁵ J

(b) 4000 J

(c) 1.82×10⁻¹⁶ J

Explanation:

Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass, v = velocity

(a)

For a moving automobile,

Ek = 1/2mv².

Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s

Substitute into equation 1

Ek = 1/2(2.0×10³)(27.78²)

Ek = 7.72×10⁵ J

(b)

For a sprinting runner,

Given: m = 80 kg, v = 10 m/s

Substitute into equation 1 above,

Ek = 1/2(80)(10²)

Ek = 40(100)

Ek = 4000 J

(c)

For a moving electron,

Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s

Substitute into equation 1 above,

Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²

Ek = 1.82×10⁻¹⁶ J

8 0
3 years ago
A freight train rolls along a track with considerable momentum. If it were to roll at the same speed but had twice as much mass,
fomenos

Answer:

The momentum would be doubled

Explanation:

The magnitude of the momentum of the freight train is given by:

p=mv

where

m is the mass of the train

v is its speed

In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:

m'=2m

therefore, the new momentum is

p'=m'v=(2m)v=2(mv)=2p

so, the momentum has also doubled.

7 0
3 years ago
A single point on a distance time graph tells the
Sonbull [250]

Answer: Instantaneous speed.

Explanation:

4 0
3 years ago
If a force of 10 N acts on an object and an additional force of 6 N acts on the object in
PIT_PIT [208]

Answer:

D

Explanation:

For this kind of problem, forces add. F = F1 + F2

F1 = 6 N

F2 = 10 N

F = 6N + 10N

F = 16N

6 0
3 years ago
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