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2 years ago

Someone help this is a test and i need to finish this quick i will give brainliest of it lets me

Physics

1 answer:

tekilochka [14]2 years ago

7 0

Sorry, I don't know but I think the correct answer is the first option.

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A 3.1 kg ball is dropped from the top of a 38 m tall building. What is the speed of the ball when it is halfway from the buildin

Archy [21]

Answer:

19.3m/s

Explanation:

Use third equation of motion

$v^2-u^2=2gh$

where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity

insert values to get answer

$v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s$

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3 years ago

What is the amount of charge when 13.5a is flowing for 2 1/2 hours

abruzzese [7]

Answer:

Quick Answer:

a. 8.9Ω

b. 1.2×104 C

Explanation:

4 0

3 years ago

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The world's most powerful laser is the LFEX laser in Japan. It can produce a 2 petawatt(2×10^15W) laser pulse that last for 1 ps

kogti [31]

Answer:

$2.82942\times 10^{24}\ W\m^2$

Explanation:

d = Diameter of spot = 30 μm

r = Radius of spot = $\frac{d}{2}=\frac{30}{2}=15\ mu m$

P = Power of the laser = $2\times 10^{15}\ W$

A = Area = $\pi r^2$

Intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{2\times 10^{15}}{\pi\times (15\times 10^{-6})^2}\\\Rightarrow I=2.82942\times 10^{24}\ W/m^2$

The light intensity within this spot is $2.82942\times 10^{24}\ W/m^2$

8 0

3 years ago

A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0

3 years ago

Which wave has a greater frequency?

DedPeter [7]

The wave diagramed in blue.

8 0

3 years ago

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