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3 years ago

9

A 1400 kg vehicle travelling at 22 m/s slows at a constant rate to 4.5 m/s in 6.75 s. What was the net force acting on the car d

uring this time?

1 answer:

Akimi4 [234]3 years ago

5 0

Answer:

Net force = 3640 N

Explanation:

From first equation of motion

V = U + at

4.5 = 22 + 6.75a

( 4.5-22) ÷ 6.75 = a

acceleration (a) = -2.6m/s^2 <em>( negative</em><em> sign shows deceleration, but actual acceleration is 2.6) </em>

But Force = mass × acceleration

Force = 1400 × 2.6

Force = 3640 N

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What are some examples of non-inertial reference frames?

Nataly_w [17]

Answer:

Examples of non-inertial reference frames

One clearcut example of an inertial reference frame is an isolated spaceship, far, far away from the Earth, the Sun, the Milky Way Galaxy, and all other massive objects. Fred places a blue ball into a claw at the left end of the ship, and red ball into a claw at the right end of the ship.

Explanation:

Let us say that you are in a car at a stop light. The car is standing still. The light turns green, and the car accelerates forward. While undergoing this acceleration, the car is a non-inertial frame of reference.

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2 years ago

The space shuttle releases a satellite into a circular orbit 630 km above the Earth.

solniwko [45]

Answer:

7,539 m/s

Explanation:

Let's use this equation to find the gravitational acceleration of this space shuttle:

$\displaystyle g=\frac{GM}{r^2}$

We know that G is the gravitational constant: 6.67 * 10^(-11) Nm²/kg².

M is the mass of the planet, which is Earth in this case: 5.972 * 10^24 kg.

r is the distance from the center of Earth to the space shuttle: radius of Earth (6.3781 * 10^6 m) + distance above the Earth (630 km → 630,000 m).

Plug these values into the equation:

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 \ Nm^2kg^-^2)(5.972\cdot 10^2^4 \ kg)}{[(6.3781\cdot 10^6 \ m)+(630000 \ m)]^2}$

Remove units to make the equation easier to read.

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 )(5.972\cdot 10^2^4 )}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Multiply the numerator out.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Add the terms in the denominator.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(7008100)]^2}$

Simplify this equation.

$\displaystyle g=8.11045189 \ \frac{m}{s^2}$

The acceleration due to gravity g = 8.11045189 m/s². Now we use the equation for acceleration for an object in circular motion which contains v and r.

$\displaystyle a = \frac{v^2}{r}$

a = g, v is the velocity that the space shuttle should be moving (what we are trying to solve for), and r is the radius we had in the previous equation when solving for g.

Plug these values into the equation and solve for v.

$\displaystyle 8.11045189 \ \frac{m}{s^2} = \frac{v^2}{7008100 \ m}$

Remove units to make the equation easier to read.

$\displaystyle 8.11045189 = \frac{v^2}{7008100}$

Multiply both sides by 7,008,100.

$56838857.89=v^2$

Take the square root of both sides.

$v=7539.154985$

The shuttle should be moving at a velocity of about 7,539 m/s when it is released into the circular orbit above Earth.

5 0

3 years ago

It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet

TEA [102]

Answer:

$v=346.05\ m.s^{-1}$

Explanation:

Given:

initial temperature of the lead bullet, $T_i=43^{\circ}C$

latent heat of fusion of lead, $L_f=2.32\times 10^4\ J.kg^{-1}$

melting point of lead, $T_m=327.3^{\circ}C$

We have:

specific heat capacity of lead, $c=129\ J.kg^{-1}.K^{-1}$

<em>According to question the whole kinetic energy gets converted into heat which establishes the relation:</em>

$\rm KE=(heat\ of\ rising\ the\ temperature\ from\ 43\ to\ 327.3\ degree\ C)+(heat\ of\ melting)$

$\frac{1}{2} m.v^2=m.c.\Delta T+m.L_f$

$\frac{1}{2} m.v^2=m(c.\Delta T+L_f)$

$\frac{v^2}{2} =129\times(327.3-43)+23200$

$v=346.05\ m.s^{-1}$

3 0

3 years ago

A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph

Olenka [21]

Answer:

B.The charge on A is -q; there is no charge on B.

Explanation:

We are given that

Charge=+q

We have to find the correct statement.

When positive charge is placed at center of uncharged metal sphere

insulated from the ground then negative charge(-q) induced on inner

surface A of sphere and the outer surface B is grounded then the surface is neutral .

It means there is no charge on surface B.

Hence, option B is true .

B.The charge on A is -q; there is no charge on B.

4 0

3 years ago

A cook removes a one-gallon pot of hot soup from a stove and places it in an ice-water bath to cool. which is the best cooling p

Verizon [17]

I thank it is #2 or #1 hop this helps;)

5 0

3 years ago