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nordsb [41]
3 years ago
10

For each of the following statements determine whether it is correct or incorrect.

Physics
1 answer:
Mrac [35]3 years ago
7 0

Answer:

a) T; b) F; c) F; d) T; e) T; f) T

Explanation:

a) True, since the equation of work done on a spring is \frac{kx^2}{2}

b) False, it will be only mgh.

c) False, since it is mv^2, no direction involved.

d) True, since the measured speed of objects depends from the reference frame of the observer.

e) True, since its formula is mv^2 and both terms are positive.

f) True, since the centripetal force the Earth feels, which is the gravitational atraction the Sun exerts, is perpendicular to its circular trajectory.

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What’s the difference between applied and pure research
Ket [755]

Answer:

Pure research is conducted without any specific goal in mind. Applied research is conducted in order to solve a specific and practical problem, unlike pure research. Hope this helped.

8 0
3 years ago
If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve o
Scilla [17]

Answer:

Maximum acceleration will be equal to 3.43m/sec^2

Explanation:

We have given coefficient of kinetic friction \mu _k=0.7

And coefficient of static friction \mu _s=1

Acceleration due to gravity g=9.8m/sec^2

When truck moves maximum force will be equal to F=\mu _kmg

It is given that half of the weight is supported by its drive wheels

So force required =\frac{\mu _kmg}{2}

From newtons law maximum acceleration will be equal to a=\frac{\frac{\mu _kmg}{2}}{m}=\frac{\mu _kg}{2}=\frac{0.7\times 9.8}{2}=3.43m/sec^2

8 0
4 years ago
A 5kg cat is lifted 2m into the air. How much GPE does it gain
luda_lava [24]

m = mass of the cat being lifted = 5 kg

h = height to which the cat is raised into air = 2 m

g = acceleration due to gravity of earth = 9.8 m/s²

U = gravitational potential energy gained by the cat

Gravitational potential energy gained by the cat is given as

U = m g h

inserting the values in the above equation

U = (5 kg) (9.8 m/s²) (2 m)

U = (5 x 9.8 x 2) (kg m²/s²)

U = 98 J

5 0
4 years ago
How do you find the angle (to the nearest tenth) on your graph that produced the greatest range, based on the line of best fit?
VikaD [51]

Answer:

yuououiy

Explanation:

3 0
3 years ago
13. A ball is thrown horizontally by a pitcher at 25.0 m/s from shoulder height (2.0m). If the catcher is 20.0 m away, will the
Doss [256]

Answer:

The ball will NOT reach the catcher before bouncing

Explanation:

y = (vertical vi)t + (1/2)gt²

2 = 0 + (1/2)(9.8)t²

t = 0.64 s

x = (horizontal vi)t + (1/2)at²

x = (25)(0.64) + 0

x = 16 m

16 m < 20 m

The ball will bounce at 16 m

7 0
3 years ago
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