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irakobra [83]
2 years ago
11

In a PE class, the teacher requires his students to monitor periodically their progress towards the fitness goals. Which of thes

e should be done by the students before and after an activity? *
a. weight monitoring
b. monitor pulse rate
c. warm up and cool down activities
d. physical check-up
Physics
1 answer:
pav-90 [236]2 years ago
8 0
The answer would be c warm up and cool down because before you do an activity stretching is very important
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What factors affect the amount of potential and kinetic energy a roller coaster has
Svetlanka [38]

Answer:

As the roller coaster goes higher, more potential energy is increased in the roller coaster. Caused by gravity and the roller coaster's position, the potential energy is stored in the roller coaster. For example, this ball is at the top of a hill, where potential energy is at it's highest. Potential energy can be calculated by Potential Energy=Mass X Acceleration X Height.

Explanation:

3 0
2 years ago
Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.
Fofino [41]

Answer:

Part a)

\alpha = 782.6 rad/s^2

Part B)

\omega = 587 rad/s

Part c)

a_t = 24.3 m/s^2

Explanation:

Part a)

As we know that

a = R \alpha

so we will have

a = 1.80 m/s^2

R = 0.230 cm

\alpha = \frac{a}{R}

\alpha = \frac{1.80}{0.230 \times 10^{-2}}

\alpha = 782.6 rad/s^2

Part B)

Angular speed of the yo-yo

\omega = \alpha t

so we have

\omega = 782.6 \times 0.750

\omega = 587 rad/s

Part c)

Tangential acceleration is given as

a_t = R \alpha

a_t = (3.10 \times 10^{-2})(782.6)

a_t = 24.3 m/s^2

6 0
3 years ago
A flying stationary kite is acted on by a force of 9.8 N downward. The wind exerts a force of 45 N at an angle of 50.0 degrees a
Savatey [412]

Answer:

38 N, 40.0° below the horizontal

Explanation:

Force exerted by an object equals mass times acceleration of that object: F = m ⨉ a. To use this formula, you need to use SI units: Newtons for force, kilograms for mass, and meters per second squared for acceleration.

7 0
2 years ago
The opening to a cave is a tall, 30-cm-wide crack. A bat that is preparing to leave the cave emits a 30 kHz ultrasonic chirp. Ho
sweet-ann [11.9K]

Answer:

Width of sound beam is 7.557 m

Explanation:

First we will calculate the wave length from given data:

λ=v/f

Were:

v is the speed

f is the frequency

\lambda=\frac{340}{30*10^3}\\ \lambda=0.01133 m

We considered the opening long and narrow, Using single slit diffraction formula:

mλ=dsinΘ

where:

d is the crack width

m is the order

Θ is angle

Considering m=1, The angle between first minimum from center of beam is:

\theta=sin^{-1}(\frac{m\lambda}{d})\\\theta=sin^{-1}(\frac{1*0.01133}{30*10^{-2}})\\ \theta=2.164^o

The width of beam is:

tanΘ=y/L

tan\theta=\frac{w/2}{L}\\ w=2L\ tan\theta\\w=2*100*tan 2.164\\w=7.557 m

Width=7.557 m

5 0
3 years ago
A small sphere is hung by a string from the ceiling of a van. When the van is stationary, the sphere hangs vertically. However,
olganol [36]

Answer with Explanation:

We are given that

String makes an angle w.r.t  vertical=\theta=

a.We have to derive an expression for the magnitude of the acceleration of the van in terms of the angle \theta and magnitude g of the acceleration due to gravity.

According to newton's second law

T sin\theta=ma

Tcos\theta=mg

\frac{Tsin\theta}{Tcos\theta}=\frac{ma}{mg}

tan\theta=\frac{a}{g}

a=gtan\theta

b.\theta=10^{\circ}

g=9.8 m/s^2

a=9.8\times tan10^{\circ}=1.73 m/s^2

c.Velocity=Constant

We have to find the angle \theta

a=0

0=9.8tan\theta

tan\theta=0

tan\theta=tan0

\theta=0^{\circ}

3 0
2 years ago
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