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4 years ago

What is the answer for this. please the correct answer only

Physics

1 answer:

omeli [17]4 years ago

5 0

Answer:

C. 900J (i tried my best)

Explanation:

the student's mass=60kg

gravitational field strength= 10N/kg:

1kg = 10N

60kg = 600N

(where 600N is the gravitational field strength of the student)

√1.5²+2² (pythagoras theorem)

=2.5m (the stairs' length)

work done= N x distance move in the direction of force (m)

=600N x 2.5m

=1500J @ Nm

1500J - 600N (gravitational field strength of the student)

=900J(??)

-correct me if i'm wrong i'm just a human :)

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The new resistance comes out to be = 4 times of original resistance .

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3 years ago

A string is vibrating between two posts as shown above. Students are to determine the speed of the wave within this string. They

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The distance between two posts

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3 years ago

A bicycle wheel with a radius of 0.42 m accelerates uniformly for 6.8 s from an initial angular velocity of 5.5 rad/s to a final

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Answer:

The angular acceleration required is 0.1765 rad/ $s^2$

Explanation:

The radius of the bicycle wheel has a radius of 0.42 m.

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Final angular velocity is given as $\omega_{f}$ = 6.7 rad/s

Therefore from the formula for angular speed we get

$\omega_{f}$ = $\omega_{0}$ + ( $\frac{d\omega}{dt}$ $\times$ t), where t is the time in seconds.

Therefore we get

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The angular acceleration required is 0.1765 rad/ $s^2$

8 0

3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

Conservation of linear momentum:

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

Conservation of kinetic energy:

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago

What is the answer for this. please the correct answer only​

What is the answer for this. please the correct answer only