The answer is <span>5.2 × 10-2 newtons. i cant do work but i know its right had the same answer in test </span>
The magnitude of normal force that floor exerts on chair is 100.83N
<u>Explanation:</u>
Given:
Weight of chair, W = 75N
Applied force, F = 42N
Angle, θ = 38°
Normal force, n = ?
We know,
Vertical component of the force = F sinθ
= 42 X sin38°
= 42 X 0.615
= 25.83N
Total normal force acting on the chair = 75N + 25.83N
= 100.83N
Therefore, the magnitude of normal force that floor exerts on chair is 100.83N
<h2><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>:-</h2>
The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave ?
<h2><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>:-</h2>
<h3>Given:-</h3>
Velocity (V) = 40 m/s
Wavelength
= 80 cm = 0.8 m
<h3>To Find:-</h3>
The frequency (F) of the wave.
<h2>Solution:-</h2>
We know,
![\bf V \: = \: F \: × \: \lambda](https://tex.z-dn.net/?f=%20%20%20%20%5Cbf%20V%20%5C%3A%20%3D%20%5C%3A%20F%20%5C%3A%20%C3%97%20%5C%3A%20%5Clambda%20)
40 = F × 0.8
F = ![\frac{40}{0.8}](https://tex.z-dn.net/?f=%20%5Cfrac%7B40%7D%7B0.8%7D%20)
F = 50
<h3>The frequency of the wave is <u>5</u><u>0</u><u> </u><u>H</u><u>z</u>. [Answer]</h3>
The magnitude of the average acceleration of the ball during this time interval will be 1969.69 m/s².
<h3>What is acceleration?</h3>
The rate of velocity change concerning time is known as acceleration.
Given data;
Initial velocity, u= 27.5 m/s
Rebound velocity, v= 21.0 m/s
Time elapsed, t = 3.30 × 10⁻³ seconds.
To find ;
Average acceleration, a
The average acceleration when the change in velocity is observed by the formula as:
![\rm a_{avg}= \frac{v-u}{t}](https://tex.z-dn.net/?f=%5Crm%20a_%7Bavg%7D%3D%20%5Cfrac%7Bv-u%7D%7Bt%7D)
Substitute the given values:
![\rm a_{avg}= \frac{27.5-21.0}{3.30 \times 10^{-3}} \\\\ a_{avg}= 1969.69 \ m/s^2](https://tex.z-dn.net/?f=%5Crm%20a_%7Bavg%7D%3D%20%5Cfrac%7B27.5-21.0%7D%7B3.30%20%5Ctimes%2010%5E%7B-3%7D%7D%20%20%5C%5C%5C%5C%20a_%7Bavg%7D%3D%201969.69%20%20%5C%20m%2Fs%5E2)
Hence, the magnitude of the average acceleration of the ball during this time interval will be 1969.69 m/s².
To learn more about acceleration, refer to the link brainly.com/question/2437624#SPJ1.
#SPJ1
Answer:
3160.41J
Explanation:
From the work energy theorem, the net work done is given by;
W = Kef - Keo
Where kef is final kinetic energy and keo is initial kinetic energy.
Now, from the question,
m = 62.8 kg
Initial speed(vo) = 5.2 m/s
Final speed (vf) = 11.3 m/s
We know that formula for kinetic energy is K = (1/2)mv²
Thus, Kef = (1/2)mvf² and Keo = (1/2)mvo²
Thus;
W = (1/2)mvf² - (1/2)mvo²
W = (m/2)(vf² - vo²)
W = (62.8/2)(11.3²-5.2²)
= 31.4(100.65) = 3160.41J