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Nitella [24]
3 years ago
7

An object can have forces acting upon it, but might not accelerate True or False?

Physics
1 answer:
vivado [14]3 years ago
8 0

Answer:

true

Explanation:

if you apply force to the top of a square it will not move

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Which occurrence demonstrates dispersion?
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A rainbow. Dispersion is the splitting of radiation into it's different wavelengths.
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The distance versus time plot for a particular object shows a quadratic relationship. Which column of distance data is possible
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The correct choice is (C)

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a rock is vertically upward with a velocity of 10 m/s. calculate the maximum height it reaches and time taken to reach that heig
lina2011 [118]

Answer:

maximum height: p(t) = Vo * t - 1/2 * g * t^2

p’(t) = v(t) = 0 = Vo - g*t. So, maximum height occurs when t = Vo / g

p(Vo / g) = Vo^2/g - 1/2 * g * (Vo/g)^2

Vo = 10 m / s. Let’s approximate g = 10 m / s^2

p(Vo / g) = 10^2 / 10 - 1/2 * 10 * (10/10)^2 = 10 - 5 = 5 meters (approximately)

Calculation of time:

v = u + gt

0 = 10√2 + (-10)t

-10√2 = -10t

2 = √2s

8 0
3 years ago
The driver of a car wishes to pass a truck that is traveling at a constant speed of 19.3 m/s . Initially, the car is also travel
miss Akunina [59]

Answer:

A)    t = 10.56 s, B)  x = 235 m, C) v = 25.2 m / s

Explanation:

A) We can solve this problem using kinematics expressions.

The distance traveled by the truck is

       x_c = v_c t

Distance traveled by the car.

The car must travel the distance that separates them from the truck x₀=25.0.   Return to the lane at x₁ = 26.5 m. the length of the truck x₂=20.7m and the length of the car x₃ = 2  4.5 = 9 m, therefore the total length traveled by the car is

          x_t = x₁ + x₂ + x₃

          x_t = 26.5 + 20.7 +9 = 56.2 m

the distance traveled by the car when it returns to the lane is

         x_c + x_t = x₀ + v₀ t + ½ a t²

when the car passes the car the distance traveled by the two vehicles is the same, we substitute

         v_c  t + x_t = x₀ + v₀ t + ½ a t²

         ½ a t² + t (v₀ -v_c) + (x₀ - x_t) = 0

we substitute the values

         ½ 0.560 t² + t (19.3 -19.3) + (25.0 - 56.2) =

         0.28 t² -31.2 = 0

         t = \sqrt{ \frac{31.2}{0.28} }

         t = 10.56 s

This is the time it takes for the car to pass the truck and back into the lane.

B) the distance traveled is

        x = v₀ t + ½ a t²

        x = 19.3 10.56 + ½ 0.560 10.56²

        x = 235 m

C) the final velocity is

         v = v₀ + a t

         v = 19.3 + 0.560 10.56

         v = 25.2 m / s

3 0
3 years ago
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