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kotykmax [81]
2 years ago
8

A rocket sled accelerates at 21.5 m/s^2 for 8.75 s. (a) What's its velocity at the end of that time? (b) How far has it traveled

?
Physics
1 answer:
Leno4ka [110]2 years ago
8 0

Answer:

(a ) vf= 188.12m/s  : Final speed at 8.75 s

(b) d= 823.04 m   : Distance the rocket sled traveled

Explanation:

Rocket sled kinematics :The rocket sled moves with a uniformly accelerated movement, then we apply the following formulas:

d =vi*t+1/2a*t² Formula (1)

vf= vi+at            Formula(2)

Where:

vi: initial speed =0

a: acceleration=21.5 m/s²

t: time=8.75 s

vf: final speed in m/s

d:displacement in meters(m)

Calculation of displacement (d) and final speed (vf)

We replace data in formulas (1) and (2):

d= 0+1/2*21.5*8.75²

d= 823.04 m

vf= 0+21.5*8.75

vf= 188.12m/s

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siniylev [52]

Answer:

The true statement is:

"(C) The magnitude of the average velocity is equal to 30 m.p.h."

Explanation:

Given that a car travels 30 miles in 1 hour on a winding mountain road.

Let' check all the statements one by one:

(A) The magnitude of the total displacement is larger than the distance traveled.

Since the entire motion of the car is not exactly given in the question, so it is not possible to tell whether the magnitude of the total displacement is larger than the distance traveled or not.

Thus, this statement is not true.

(B) The magnitude of the average velocity is greater than 30 m.p.h.

The average velocity of an object is defined as the total displacement covered by the particle divided by the total time taken in covering that displacement.

Total distance covered by the car = 30 miles.

Total time taken by the car to cover this distance = 1 hour.

Therefore, the average velocity of the car for this time interval = \rm \dfrac{30\ miles}{1\ hour }= 30\ m.p.h.

Thus, this statement is also not true.

(C) The magnitude of the average velocity is equal to 30 m.p.h.

As is cleared in part (B) section above, the average velocity of the car in the given time interval is 30 m.p.h.

Thus, this statement is true.

(D)The magnitude of the average velocity is less than to 30 m.p.h.

Since. the average velocity of the car is 30 m.p.h.

Thus, this statement is not true.

(E)The car traveled with a constant speed of 30 m.p.h.

The motion of the car on the mountain road is not thoroughly given in the question, so again it is not possible to tell whether the car traveled with a constant speed of 30 m.p.h. or not.

Thus, this statement is also not true.

4 0
2 years ago
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A beam of monochromatic light (f =5.09 ×1014 Hz) has a wavelength of 589 nanometers in air. What is the wavelength of this light
frosja888 [35]
Lucite has a refractive index of n=1.50. This means that the speed of the light in lucite is decreased according to:
v=\frac{c}{n}
where c=3 \cdot 10^8 m/s is the speed of light in air. Putting the number in the formula, we find that the speed of light in lucite is
v=\frac{3 \cdot 10^8 m/s}{1.50}=2\cdot 10^8 m/s
The frequency of the light is f=5.09 \cdot 10^{14}Hz, so now we can calculate the wavelength in lucite by using the formula:
\lambda=\frac{v}{f}=\frac{2\cdot 10^8 m/s}{5.09 \cdot 10^{14} Hz}=3.93 \cdot 10^{-7} m=393 nm
<span>Therefore, the correct answer is (2) 393 nm.</span>
7 0
2 years ago
Which of the following is the BEST example of increasing the intensity of a workout? A. running one mile further than normal B.
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Answer:

B. running one mile faster than normal

Explanation:

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2 years ago
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A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s2 . How big is the frictional force on the block
solniwko [45]
We can first calculate the net force using the given information.

By Newton's second law, F(net) = ma:

F(net) = 25 * 4.3 = 107.5

We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):

f = F(net) - F(app) = 150 - 107.5 = 42.5 N

Now we can calculate the coefficient of friction, u, using the normal force, F(N):

f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
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2 years ago
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im
geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

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          Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

         

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

8 0
2 years ago
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