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3 years ago

How do you determine how much kinetic energy an object has

Physics

2 answers:

IgorC [24]3 years ago

7 0

Answer:

the answer is a

hope it helps

Ierofanga [76]3 years ago

3 0

I believe the answer is a. Because the formula of kinetic energy is 1/2(m)•(v^2)

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A person sitting still in a moving airplane is considered to be... (SP1a) a. Motionless in relation to everything in the plane.

vichka [17]

Answer:

C. Both a and b

Explanation:

Firstly, persons and objects in a moving plane as described in this question, are moving at the same speed as the plane even if there is no individual movement of these objects.

However, this question describes a person sitting still in a moving plane. This means that;

- The person is motionless in relation to everything in the plane i.e the person is not moving even if other things in the plane are.

- The person is in motion compared to everything on the ground i.e. the person is moving at the same speed as the plane, hence, in comparison with the ground, the person is moving.

Therefore, options A and B are correct

6 0

3 years ago

A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo

BARSIC [14]

We are given that,

$\frac{dx}{dt} = 4ft/s$

We need to find $\frac{d\theta}{dt}$ when $x=8ft$

The equation that relates x and $\theta$ can be written as,

$\frac{x}{6} tan\theta$

$x = 6tan\theta$

Differentiating each side with respect to t, we get,

$\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$

$\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}$

$\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}$

Replacing the value of the velocity

$\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2$

$\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta$

The value of $cos \theta$ could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

$cos\theta = \frac{6}{10}$

$\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2$

$\frac{d\theta}{dt} = \frac{24}{25}$

Search light is rotating at a rate of 0.96rad/s

4 0

3 years ago

A cat is being chased by a dog. Both are running in a straight line at constant speeds. The cat has a head start of 3.8 m. The

Mars2501 [29]

In 6 secs, the dog covers-
S=vt
8.9*6 = 53.4 m.
In the same time, the cat covers, 53.4-3.8 = 49.6 m.
Thus, speed of the cat, v= s/t,
= 49.6/6 = 8.267 m/s

7 0

4 years ago

a mountain biker cycling a trail traveling at 5.4 m/s begins to slow down when she notice a wooden bridge ahead has flooded. if

andrew-mc [135]

Answer:

t = 5.56 s

Explanation:

In order to calculate the time interval taken by the mountain biker to come to a stop, we will use third equation of motion and first find the deceleration:

2as = Vf² - Vi²

where,

a = deceleration = ?

s = distance = 15 m

Vf = Final Velocity = 0 m/s

Vi = Initial Velocity = 5.4 m/s

Therefore,

2a(15 m) = (0 m/s²) - (5.4 m/s)²

a = - 0.972 m/s²

Now, we use 1st equation of motion:

Vf = Vi + at

therefore,

0 m/s = 5.4 m/s + (-0.972 m/s²)(t)

t = (5.4 m/s)/(0.972 m/s²)

5 0

3 years ago

A street light is on top of a 9 foot pole. Joe, who is 3 feet tall, walks away from the pole at a rate of 4 feet per second. At

Gekata [30.6K]

Answer:2 ft/s

Explanation:

Given

Length of Pole is 9 ft

Length of Joe is 3 ft

Joe walks away from Pole at the rate 4 ft/s

Let Joe is x m away from Pole so its shadow length is x'

From Similar triangle concept

$\frac{x'}{x+x'}=\frac{3}{9}$

3x'=x+x'

x=2x'

and it is given $\frac{\mathrm{d} x}{\mathrm{d} t}=4 ft/s$

Differentiating

$\frac{\mathrm{d} x}{\mathrm{d} t}=2\frac{\mathrm{d} x'}{\mathrm{d} t}$

$4=2\times \frac{\mathrm{d} x'}{\mathrm{d} t}$

$\frac{\mathrm{d} x'}{\mathrm{d} t}=2 ft/s$

6 0

3 years ago