This her sisterrrrrrrrrrrrr

A man standing on a bus remains still when the bus is at rest. When the bus moves forward and then slows down the man continues

Stells [14]

C. inertia. the man is sent flying off the bus because of his weight and the sudden stop of the bus. this effect is called inertia. an example of gravity would be throwing an apple up and having it come to the ground. an example of weight would be putting a man and an elephant on a scale and having the elephant come down while the man goes up.

6 0

3 years ago

Consider electromagnetic waves in free space. What is the wavelength of a wave that has the following frequencies? (a) 4.10 x 10

navik [9.2K]

Explanation:

To find the answer use the equation speed of light=wavelength multiplied by frequency (c=lambda*f) by substituting the value for the frequency the the speed of light

7 0

3 years ago

A baseball player friend of yours wants to determine his pitching speed. you have him stand on a ledge and throw the ball horizo

zhenek [66]

Answer:

The pitching speed of your friend is 33.20 m/s

Explanation:

Lets explain how to solve the problem

Your friend throw the ball horizontally that means the vertical initial

component of velocity is zero ( $u_{y}=0$ ).

The ball is thrown from a height 4 meters above the ground.

The height $h=u_{y}t+\frac{1}{2}gt^{2}$

Remember: the height is negative value because its below the point of

thrown (initial position)

h = -4 m , $u_{y}=0$ and g = -9.8 m/s²(downward)

Substitute these values in the rule above

⇒ $4=0-\frac{1}{2}(9.8)t^{2}$

⇒ -4 = -4.9t² (multiply both sides by -1)

⇒ 4 = 4.9t² (divide both sides by 4.9)

⇒ 0.81633 = t² (take √ for both sides)

⇒ t = 0.9035

Then the time of the ball to land on the ground is 0.9035 seconds

The range of the ball on the ground is 30 m

The range $R=u_{x}*t$ , where $u_{x}$ is the horizontal

component of the initial velocity

R = 30 meters and t = 0.9035

⇒ $30=u_{x}(0.9035)$ (divide both sides by 0.9035)

⇒ $u_{x}=33.20$ m/s

The pitching speed of your friend is 33.20 m/s 

4 0

3 years ago

A ball thrown horizontally at vi = 30.0 m/s travels a horizontal distance of d = 55.0 m before hitting the ground. from what hei

tekilochka [14]

Assume no air resistance, and g = 9.8 m/s².

Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity

The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s

With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0

Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°

The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s

If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)

Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s

The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m

Answer: h = 110.4 m

7 0

3 years ago

When 1,250^3/4 is written in simplest radical form, which value remains under the radical?

GaryK [48]

Answer:

$125\sqrt[4]{8}$