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3 years ago

1. How many hours in 9.00 years?

Physics

1 answer:

Rudiy273 years ago

8 0

Answer:

78840

Explanation:

Google gets the answer quicker btw

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erastovalidia [21]

Answer:11.5m

Explanation:

6 0

2 years ago

HELP ME PLEASE WITH MY HOMEWORK ILL GIVE BRAINLIEST

nadezda [96]

Answer: Mass

Explanation:

8 0

2 years ago

A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten

kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

$\mu$ = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

$\omega$ = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

$P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W$

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

7 0

3 years ago

Block A is accelerating with Block B at a rate of 0.800 m/s2 along a frictionless surface. It suddenly encounters a surface that

Natalija [7]

Answer:

$a=0.6\ m/s^2$

Explanation:

The attached figure shows the whole description. Considering the applied force is 100 N.

The acceleration of both blocks A and B, $a=0.8\ m/s^2$

Firstly calculating the mass m using the second law of motion as :

F = ma

m is the mass

$m=\dfrac{F}{a}$

$m=\dfrac{100\ N}{0.8\ m/s^2}$

m = 125 kg

It suddenly encounters a surface that supplies 25.0 N a friction, F' = 25 N

$(F-F')=ma$

$(100-25)=125\times a$

$a=\dfrac{75}{125}=0.6\ m/s^2$

So, the new acceleration of the block is $0.6\ m/s^2$ . Hence, this is the required solution.

7 0

2 years ago

A cannonball is launched from the ground at an angle of 30 degrees above the horizontal and a speed of 30 m/s. Ideally (no air r

marusya05 [52]

As we know that here no air resistance while ball is moving in air

So here we will say that

initial total energy = final total energy

$KE_i + U_i = KE_f + U_f$

here we know that

$Ui = U_f = 0$ (as it will be on ground at initial and final position)

so we will say

$KE_i = KE_f$

since mass is always conserved

so we will say that final speed of the ball must be equal to the initial speed of the ball

so we have

$v_f = v_i = 30 m/s$

6 0

2 years ago

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