Mr. Phillips' car iso parked on a steep hill

A truck covers 40.0 m in 7.45 s while uniformly slowing down to a final velocity of 3.50 m/s.

astra-53 [7]

Answer:

(A) Original velocity will be 7.244 m /sec

(B) Acceleration will be $-0.5026m/sec^2$

Explanation:

We have given distance covers by truck s = 40 m

Time taken by truck to cover this distance t = 7.45 sec

Final velocity v = 3.50 sec

According to second equation of motion

$S=ut+\frac{1}{2}at^2$

$40=u\times 7.45+\frac{1}{2}\times a\times 7.45^2$

$7.45u+27.751a=40$ -----eqn 1

According to first equation of motion

v = u + at

So $3.5=u+7.45a$ -----eqn2

Solving equation 1 and 2

a = $-0.5026m/sec^2$

And u = 7.244 m /sec

(A) Original velocity will be 7.244 m /sec

(B) Acceleration will be

$-0.5026m/sec^2$

6 0

4 years ago

A bird is flying with a speed of 18.0 m/s over water when it accidentally drops a 2.00 kg fish. If the altitude of the bird is 5

Alina [70]

Here we go.
My abbreviations; KE = Kinetic Energy; GPE = Gravitational Potential Energy.

So first off, we know the fish has KE right when the bird releases it. Why? Because it has horizontal velocity after released! So let’s calculate it:
KE = 1/2(m)(V)^2
KE = 1/2(2)(18)^2
KE = 324 J

Nice!
We also know that the fish has GPE at its maximum height before release:
GPE = mgh
GPE = (2)(9.81)(5.40)
GPE = 105.95 J

Now, based on the *queue dramatic voice* LAW OF CONSERVATION OF ENERGY, we know all of the initial energy of the fish will be equal to the amount of final energy. And since the only form of energy when it hits the water is KE, we can write:
KEi + GPEi = KEf
(Remember - we found the initial energies before!)
(324) + (105.95) = KEf
KEf = 429.95J
And that’s you’re final answer! Notice how this value is MORE than the initial KE from before (324 J) - this is because all of the initial GPE from before was transformed into more KE as the fish fell (h decreased) and sped up (V increased).

If this helped please like it and comment!

4 0

3 years ago

A car drives over a hilltop that has a radius of curvature 0.120 km at the top of the hill. At what speed would the car be trave

Mashutka [201]

Answer:

34.3 m/s

Explanation:

Newton's Second Law states that the resultant of the forces acting on the car is equal to the product between the mass of the car, m, and the centripetal acceleration $a_c$ (because the car is moving of circular motion). So at the top of the hill the equation of the forces is:

$mg-R = m a_c = m\frac{v^2}{r}$

where

(mg) is the weight of the car (downward), with m being the car's mass and g=9.8 m/s^2 is the acceleration due to gravity

R is the normal reaction exerted by the road on the car (upward, so with negative sign)

v is the speed of the car

r = 0.120 km = 120 m is the radius of the curve

The problem is asking for the speed that the car would have when it tires just barely lose contact with the road: this means requiring that the normal reaction is zero, R=0. Substituting into the equation and solving for v, we find:

$v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(120 m)}=34.3 m/s$

6 0

3 years ago

Mg represents the element

Romashka-Z-Leto [24]

Answer:

Mg is the symbol for Magnesium. Atomic #- 12

Explanation:

6 0

4 years ago

Which statement is correct about the car?

JulijaS [17]

Answer:

B

Explanation:

OOf we are doing this stuff atm

So if its faster at the front and slow at the back you can tell that its not slowing down because less of a force is there however at the front there is more of a force. Friction is low which means that its not makimg much contact so no sudden change of forces thats also why its B

6 0

3 years ago

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