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Vlad [161]
3 years ago
11

An energetic child runs forward and backward. It's motion is shown on the following graph of horizontal position x vs time t

Physics
1 answer:
Klio2033 [76]3 years ago
6 0

Answer: 0.50 m/s

Explanation: it’s on khan academy this was the correct answer

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A chair of weight 70.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F =
Annette [7]

Answer:

94.67 N

Explanation:

Consider a free body diagram with force, F of 41 N applied at an angle of 37 degrees while the weight acts downwards. Resolving the force into vertical and horizontal components, we obtain a free body diagram attached.

At equilibrium, normal reaction is equal to the sum of the weight and the vertical component of the force applied. Applying the condition of equilibrium along the vertical direction.  

\begin{array}{l}\\\Sigma {F_y} = 0\\\\N - W - F\sin \theta = 0\\\\N = W + F\sin \theta \\\end{array}

Substituting 70 N for W, 41 N for F and \theta for 37 degrees

N=70+41sin37=94.67441595  N and rounding off to 2 decimal places

N=94.67 N

​

6 0
4 years ago
Which of the following devices could be used to create a position--time graph for an object? 
pickupchik [31]
D none of the above

all of them is related to energy and motion , not to positioning.
6 0
4 years ago
A hockey puck slides off the edge of a platform with an initial velocity of 20 m/s horizontally. The height of the platform abov
Rina8888 [55]

Answer:

20.96 m/s

Explanation:

Using the equations of motion

y = uᵧt + gt²/2

Since the puck slides off horizontally,

uᵧ = vertical component of the initial velocity of the puck = 0 m/s

y = vertical height of the platform = 2 m

g = 9.8 m/s²

t = time of flight of the puck = ?

2 = (0)(t) + 9.8 t²/2

4.9t² = 2

t = 0.639 s

For the horizontal component of the motion

x = uₓt + gt²/2

x = horizontal distance covered by the puck

uₓ = horizontal component of the initial velocity = 20 m/s

g = 0 m/s² as there's no acceleration component in the x-direction

t = 0.639 s

x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m

For the final velocity, we'll calculate the horizontal and vertical components

vₓ² = uₓ² + 2gx

g = 0 m/s²

vₓ = uₓ = 20 m/s

Vertical component

vᵧ² = uᵧ² + 2gy

vᵧ² = 0 + 2×9.8×2

vᵧ = 6.26 m/s

vₓ = 20 m/s, vᵧ = 6.26 m/s

Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s

4 0
3 years ago
Read 2 more answers
What is the study of the mone and stars
maks197457 [2]
The study of the moon as well as the stars is astronomy
8 0
3 years ago
The latitude of any location on earth is the angle formed by the two rays drawn from the center of earth to the location and to
Alina [70]

The distance between city a and city b is 833.345 miles.

We know that

1°=60'

The distance of city a from the initial ray is  calculated as

x_a=3960*tan45.46°=4024.101 miles

The distance of city b from the initial ray is calculated as

x_b=3960*tan 38.86°=3190.75 miles

Now the distance between city a and b is equal to

4024.101-3190.75=833.345 miles

This is the vertical distance between the cities.

5 0
3 years ago
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