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topjm [15]
3 years ago
9

A car that is standing still accelerates up a hill with a slope of 6.4% at an average acceleration of 2.93 ft/s^2. The hill is 1

2.42 miles long, and the final velocity of the car is 203 mph. How long does it take for the car to climb to the top of the hill?
Physics
1 answer:
alexgriva [62]3 years ago
5 0

Answer:

213 s

Explanation:

Slope is the ratio of change in vertical distance to change in horizontal distance.

Slope = vertical height / horizontal height

Therefore:

6.4% = vertical height / 12.42

vertical height = 6.4% * 12.42

vertical height = 0.8 miles

The distance travelled by the car (s) is:

s² = 0.8² + 12.42²

s² = 154.9

s = 12.45 miles

Acceleration (a) =  2.93 ft/s^2 = 0.00055 mile/s²

initial velocity (u) = 0, final velocity = 203 mph

Using:

s = ut + 0.5at²

12.45 = 0.5(0.00055)t²

t =213 s

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3 years ago
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A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c
den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, L_1=L-1

New time period of the pendulum is, T_1=2.81\ s

We know that, the time period of a simple pendulum of length 'L' is given as:

T=2\pi\sqrt{\frac{L}{g}}-------------- (1)

So, for the new length, the time period is given as:

T_1=2\pi\sqrt{\frac{L_1}{g}}------------ (2)

Squaring both the equations and then dividing them, we get:

\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1

Now, plug in the given values and calculate 'L'. This gives,

L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m

Therefore, the original length of the simple pendulum is 2.97 m

4 0
3 years ago
A 4350 kg truck, driving 7.39 m/s, runs into the back of a stationary car. After the collision, the truck moves 4.55 m/s and the
DiKsa [7]

Answer:

Mass of car = 1098 kg

Explanation:

Here law of conservation of momentum is applied.

Let mass of car be m.

Initial momentum = Final momentum.

Initial momentum = 4350 x 7.39 + m x 0 = 32416.5 kgm/s

Final momentum = 4350 x 4.55 + m x 11.5 = 19792.5+11.5m

We have

      19792.5+11.5m = 32416.5

        m = 1097.97 kg

Mass of car = 1098 kg

4 0
3 years ago
If an object is dropped from a height of meters, the velocity (in ) at impact is given by v= √2gh , where g=9.8m/sec^2 is the ac
Neporo4naja [7]

Answer:

A.) V = 14 m/s

B.) h = 36.6 m

Explanation:

Given the formula v = √2gh

where g = 9.8m/sec^2 is the acceleration due to gravity.

A.) Determine the impact velocity for an object dropped from a height of 10 m.

Substitute height h in the given formula

V = √2gh 

V = √2 × 9.8 × 10

V = √196

V = 14 m/s

b. Determine the height required for an object to have an impact velocity of 26.8 m/sec (~ 60 mph). Round to the nearest tenth of a meter.

Substitute the velocity in the given formula and make height h the subject of formula.

26.8 = √2 × 9.8 × h

Square both sides

718.24 = 19.6h

h = 718.24 / 19.6

h = 36.64 m

h = 36.6 m

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3 years ago
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luda_lava [24]
Hmmm the only formula i know is v = d * t  or volocity = distance * time
5 0
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