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3 years ago

4. A 75 kg bobsled is pushed along a horizontal surface by two athletes. After the

Physics

1 answer:

Alenkinab [10]3 years ago

8 0

The net force on the sled is 300 N

Explanation:

First of all, we start by finding the acceleration of the bobsled, by using the suvat equation:

$v^2-u^2=2as$

where:

v = 6.0 m/s is the final velocity of the sled

u = 0 is the initial velocity

a is the acceleration

s = 4.5 m is the displacement of the sled

Solving for a, we find

$a=\frac{v^2-u^2}{2s}=\frac{6.0^2-0}{2(4.5)}=4 m/s^2$

Now we can find the net force on the sled by using Newton's second law:

F = ma

where

F is the net force

m = 75 kg is the mass of the sled

$a=4 m/s^2$ is the acceleration

Solving the equation, we find the net force:

$F=(75)(4)=300 N$

Learn more about acceleration and Newton laws here:

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V125BC [204]

Answer:

Current in the toroid will be $I=17.32\times 10^{-3}A$

Explanation:

We have given number of winding in rectangular toroid N = 1500

Self inductance of toroid L = 0.06 H

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We have to find the current in the toroid

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$9\times 10^{-6}=\frac{1}{2}\times 0.06\times I^2$

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So current in the toroid will be $I=17.32\times 10^{-3}A$

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4 years ago

Use the following equation to help you answer the question. The peak intensity of radiation from a star named Sigma is 2 x 10 6

puteri [66]

Answer:

1449 K

Explanation:

The surface temperature of a star is related to its peak wavelength by Wien's displacement law:

$T=\frac{b}{\lambda}$

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T is the surface temperature

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3 years ago

Read 2 more answers

igor_vitrenko [27]

Answer:

$a=40\ m/s^2$

Explanation:

Given that,

Initial speed of a shuttlecock, u = 30 m/s

Final speed of the shuttlecock, v = 10 m/s

Time, t = 0.5 s

We need to find its average acceleration. The acceleration of an object is equal to the change in speed divided by time taken. It is given by :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{10-30}{0.5}\\\\a=-40\ m/s^2$

So, the average acceleration of badminton shuttlecock is $40\ m/s^2$ .

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3 years ago

A spring-loaded gun can fire a projectile to a height h if it is fired straight up. if the same gun is pointed at an angle of 45

Naily [24]

When firing straight up:
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Then,
0 = u^2 - 2gh
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When firing at 45°,
Initial velocity, U = u Sin 45 = Sqrt (2gh)·Sin 45

Maximum height, H = U^2*(Sin Ф)^2/2g

substituting;
H = [Sqrt (2gh)·Sin 45]^2*(Sin 45)^2]/2g
H = [2gh*(Sin 45)^2*(Sin 45)^2]/2g
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4 years ago

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Veronika [31]

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3 years ago