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Black_prince [1.1K]

3 years ago

13

Velocity of a car that travled 120 miles in 2 hours

2 answers:

Sloan [31]3 years ago

5 0

My answer to your question is 60 mph 60×2 = 120

Sophie [7]3 years ago

4 0

60 Miles per hour 60 times 2 is 120

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A man standing in front of a mountain beats a drum at regular intervals. The rate of drumming is generally increased and he find

Likurg_2 [28]

I think it is d I hope this help you if not let me know if it is not right

3 0

3 years ago

In a physics lab, light with a wavelength of 530 nm travels in air from a laser to a photocell in a time of 16.7 ns . When a sla

Harlamova29_29 [7]

Answer:

$\lambda'=78.086\ nm$

Explanation:

Given:

wavelength of light in the air, $\lambda=530\times 10^{-9}\ m$

time taken to travel from the source to the photocell via air, $t=16.7\ s$

time taken to reach the photocell via air and glass slab, $t'=21.3\times 10^{-9}\ s$

thickness of the glass slab, $x=0.87\ m$

<u>Now we have the relation for time:</u>

$\rm time=\frac{distance}{speed}$

hence,

$t=\frac{d}{c}$

c= speed of light in air

$16.7\times 10^{-9}=\frac{d}{3\times 10^8}$

$d=16.7\times 10^{-9}\times 3\times 10^8$

$d=5.01\ m$

For the case when glass slab is inserted between the path of light:

$\frac{(d-x)}{c} +\frac{x}{v} =t'$ (since light travel with the speed c only in the air)

here:

v = speed of light in the glass

$\frac{(5.01-0.87)}{3\times 10^8} +\frac{0.87}{v} =21.3\times 10^{-9}$

$v=4.42\times 10^7\ m.s^{-1}$

Using Snell's law:

$\frac{\lambda}{\lambda'} =\frac{c}{v}$

$\frac{530}{\lambda'} =\frac{3\times 10^8}{4.42\times 10^7}$

$\lambda'=78.086\ nm$

5 0

3 years ago

In a fireworks display, a rocket is launched from the ground with a speed of 18.0 m/s and a direction of 51.0° above the horizon

Masja [62]

Answer:38.66 m

Explanation:

Given

launch angle $\theta =51^{\circ}$

launch velocity $u=18 m/s$

center of mass continue to travel its original Path so it center of mass will be at a distance of

$R=\frac{u^2\sin 2\theta }{g}$

$R=\frac{18^2\sin 102}{9.8}$

$R=32.33 m$

Center of mass will be at $x=32.33 m$

(b)if one of the piece will be at x=26 m then other will be at

$x_{com}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

$x_{com}=\frac{\frac{m}{2}\cdot 26+\frac{m}{2}\cdot x_0}{\frac{m}{2}+\frac{m}{2}}$

$32.33=\frac{26+x_0}{2}$

$x_0=38.66 m$

8 0

3 years ago

A 129-kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 65.

LUCKY_DIMON [66]

Answer:

Moment of inertia of the system is 289.088 kg.m^2

Explanation:

Given:

Mass of the platform which is a uniform disk = 129 kg

Radius of the disk rotating about vertical axis = 1.61 m

Mass of the person standing on platform = 65.7 kg

Distance from the center of platform = 1.07 m

Mass of the dog on the platform = 27.3 kg

Distance from center of platform = 1.31 m

We have to calculate the moment of inertia.

Formula:

MOI of disk = $\frac{MR^2}{2}$

Moment of inertia of the person and the dog will be mr^2.

Where m and r are different for both the bodies.

So,

Moment of inertia $(I_y_y )$ of the system with respect to the axis yy.

⇒ $I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g$

⇒ $I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2$

⇒ $I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2$

⇒ $I_y_y=289.088\ kg.m^2$

The moment of inertia of the system is 289.088 kg.m^2

7 0

3 years ago

What does the term calorie refer to

OLga [1]

Hey there,
<span>It means- the amount of energy required to raise the temperature of a liter of water one degree centigrade at sea level
</span>
Hope this helps :))

~Top

6 0

3 years ago