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Elden [556K]
4 years ago
11

You should always wear your seat belt just in case the car comes to an abrupt stop. The seat belt will hold you in place so that

your body does not continue moving when the car stops.
Physics
2 answers:
Schach [20]4 years ago
7 0

Answer:

Inertia, first

Explanation:

did the assignment

Gwar [14]4 years ago
5 0
Newtons 1st law of motion states that the object will continue to move at its present speed and direction until an outside force acts upon it.
 
So unless the objects inside the car are restrained, they will continue moving at whatever speed the car is traveling at, even if the car is stopped by a crash.

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Mimas, a moon of Saturn, has an orbital radius of 1.62 × 108 m and an orbital period of about 23.21 h. Use Newton’s version of K
Drupady [299]

Answer:

3.60432\times 10^{26}\ kg

Explanation:

a = Orbital radius = 1.62\times 10^8\ m

T = Orbital period = 23.21 hours

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

From Kepler's third law we get

M=\frac{4\pi^2a^3}{GT^2}\\\Rightarrow M=\frac{4\pi^2\times (1.62\times 10^8)^3}{6.67\times 10^{-11}\times (23.21\times 3600)^2}\\\Rightarrow M=3.60432\times 10^{26}\ kg

From the given data the mass of Saturn is 3.60432\times 10^{26}\ kg

8 0
3 years ago
A railroad car of mass M moving at a speed v1 collides and couples with two coupled railroad cars, each of the same mass M and m
marshall27 [118]

Answer:

vf = v₁/3 + 2v₂/3

Explanation:

Using the law of conservation of linear momentum,

momentum before impact = momentum after impact

So, Mv₁ + 2Mv₂ = 3Mv (since the railroad cars combine) where v₁ = initial velocity of first railroad car, v₂ = initial velocity of the other two coupled railroad cars, and vf = final velocity of the three railroad cars after impact.

Mv₁ + 2Mv₂ = 3Mvf

dividing through by 3M, we have

v₁/3 + 2v₂/3 = vf

vf = v₁/3 + 2v₂/3

6 0
3 years ago
Which part of a sandwich is most like a homogeneous mixture?
hoa [83]

Bread is considered to be a heterogeneous mixture. It is a heterogeneous mixture because all of the components that are used to make the bread are physically separate.

6 0
3 years ago
Read 2 more answers
If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric
poizon [28]

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

3 0
3 years ago
A car starting from rest moves with a constant acceleration of 10 mi/hr2 for 1 hour, then decelerates at a constant 5 mi/hr2 unt
Vinvika [58]

Answer:

The total distance traveled by the car (S) = 15 mi

Explanation:

Total acceleration (a_{1}) = 10 mi / h^{2}

Time (t) = 1 hour

Deceleration (a_{2}) = - 5 mi / h^{2}

From law of speed we know that  V = u + a_{1} t -----------(1)

Where V = Final speed

           u = initial speed and t = time period

Car starts from rest so initial speed is zero so u = 0

Then equation 1 becomes V = a_{1} t = 10 × 1 = 10 mi / h ---------(2)

The distance traveled by the car S_{1} = u t + \frac{1}{2} × a_{1} × t^{2} -------------(3)

Put all the values in equation we get S_{1} = 0 + \frac{1}{2} × 10 × 1^{2}

                                                             S_{1} = 5 mi ---------------(4)

This is the distance traveled by the car in 1 hour.

In second case when car starts decelerates the equation of motion becomes

⇒ V = u - a t -----------(5)

final speed in this case = 0 and initial speed (u) = 10 mi / h

⇒ 0 = 10 - 5 t

⇒ 5 t = 10

⇒ t = 2 hours

Distance traveled by the car in to hours  S_{2} = u t + \frac{1}{2} × a_{2} × t^{2}

⇒ S_{2} = 10 × 2 - \frac{1}{2} × 5 × 2^{2}

⇒ S_{2} = 20 - 10

⇒ S_{2} = 10 mi

This is the distance traveled by the car in the next two hours.

Thus the total distance traveled by the car = S_{1} + S_{2}

⇒ S = 5 + 10 = 15 mi

3 0
3 years ago
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