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3 years ago

Can someone help me with this whole page? i can't seem to figure it out. Is physics homework

Physics

1 answer:

Roman55 [17]3 years ago

7 0

Go on khan academy they help you with all of this stuff and you will know more

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A pitcher throws a 0.145-kg baseball at a velocity of 30.0m/s. How

nata0808 [166]

Answer: 65.25 J

Explanation:

Kinetic Energy K.E. = 1/2 * m * v^2 ; where m is the mass of the body and v is the velocity of the body ; K.E. = 1/2 * 0.145 * 30 * 30 = 65.25 Joules

6 0

3 years ago

What happens to kinetic energy when you increase the mass?

Rus_ich [418]

Increasing mass increases kinetic energy. This can be seen in the equation KE = 1/2 (m) (v)^2

If you found this helpful, please brainliest me!

5 0

3 years ago

In traveling a distance of 2.5 km between points A and D, a car is driven at 99 km/h from A to B for t seconds and 48 km/h from

Andreyy89

Answer:

d = 1.954 Km

Explanation:

given,

total distance, D = 2.5 Km

in stretch A to B =

speed = 99 Km/h = 99 x 0.278 = 27.22 m/s time =t

in stretch B to C

time = 3.4 s

In stretch C to D

speed = 48 Km/h = 48 x 0.278 = 13.34 m/s time =t

we know,

distance = speed x time

distance of BC

using equation of motion

v = u + a t

27.22 = 13.34 - a x 3.4

a = 4.08 m/s²

uniform deceleration is equal to 4.08 m/s²

distance traveled in BC

$s = ut + \dfrac{1}{2}at^2$

$s = 13.34\times 3.4 + \dfrac{1}{2}\times 4.08 \times 3.4^2$

s = 68.94 m

$3000 = 99 \times \dfrac{1000\ t}{3600}+ 68.94 + 48\times \dfrac{1000\ t}{3600}$

3000 = 27.5 t + 68.94 + 13.33 t

40.83 t = 2931.06

t = 71.79 s

distance travel in AB

distance = s x t

d = 27.22 x 71.79

d = 1954 m

d = 1.954 Km

distance between A and B is equal to 1.954 Km.

4 0

3 years ago

One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe

Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

$\mu = \frac{m}{l}$

$\mu = \frac{8.79*10^{-3}}{70*10^{-2}}$

$\mu = 0.01255kg/m$

The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

Replacing we have

$\lambda = \frac{2}{3} (70*10^{-2})$

$\lambda = 0.466m$

The frequency of the sound wave is

$f_s = \frac{v}{\lambda_s}$

$f_s = \frac{344}{0.768}$

$f_s = 448Hz$

Now the velocity of the wave would be

$v = f_s \lambda$

$v = (448)(0.466)$

$v = 208.768m/s$

The expression that relates the velocity of the wave, tension on the string and linear mass density is

$v = \sqrt{\frac{T}{\mu}}$

$v^2 = \frac{T}{\mu}$

$T= \mu v^2$

$T = (0.01255kg/m)(208.768m/s)^2$

$T = 547N$

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the $n^{th}$ harmonic frequency is

$f_n = nf_1$

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

$n=3$