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4 years ago

9

You push a block to the top of an inclined plane. The mechanical advantage of the plane is 5. The inclined plane is 10 meters lo

ng. What is the height of the inclined plane?
10 m
50 m
2 m
0.5 m

2 answers:

Veronika [31]4 years ago

4 0

Answer: 50 m

Explanation:

eduard4 years ago

4 0

Answer:

the answer is 2m

Explanation:

the formula to find the mechanical advantage of an inclined plane is

MA (mechanical advantage) = L (length of plane) / H (height)

$MA = \frac{L}{H}$

for this problem we already know the MA and length. simply plug that into the formula and isolate H.... you'll end up finding the height >> 2m

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A glider is gliding through the air at a height of 416 meters with a speed of 45.2 m/s. The glider

julia-pushkina [17]

There's not enough information to find an answer.

I think the idea here is that in descending (416 - 278) = 138 meters,
the glider gives up some gravitational potential energy, which
becomes kinetic energy at the lower altitude. This is all well and
good, but we can't calculate the difference in potential energy
without knowing the mass of the glider.

3 0

3 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 1

Bumek [7]

Answer:

202.8m

Explanation:

Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.

First calculate the total time travelled by using the second equation of motion

h = Ut + 1/2gt^2

Let assume that u = 0

And h = 3.5

Substitute all the parameters into the formula

3.5 = 1/2 × 9.8 × t^2

3.5 = 4.9t^2

t^2 = 3.5/4.9

t^2 = 0.7

t = 0.845s

To know how far the cannonball travel, let's use the equation

S = UT + 1/2at^2

But acceleration a = 0

T = 2t

T = 1.69s

S = 120 × 1.69

S = 202.834 m

Therefore, the distance travelled by the cannon ball is approximately 202.8m.

4 0

3 years ago

A 16.0kg canoe moving to the left at 12.5m/s makes an elastic head-on collision with a 14.0kg raft moving to the right at 16.0m/

kherson [118]

The canoe is moving at 14.1 m/s to the right after the collision.

Explanation:

According to the law of conservation of momentum, in absence of external forces the total momentum of the system must be conserved before and after the collision. So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 16.0 kg$ is the mass of the canoe

$u_1 = -12.5 m/s$ is the initial velocity of canoe (we take right as positive direction, and since the canoe is moving to the left, its velocity is negative)

$v_1$ is the final velocity of the canoe

$m_2 = 14.0 kg$ is the mass of the raft

$u_2 = +16.0 m/s$ is the initial velocity of the raft

$v_2 = -14.4 m/s$ is the final velocity of the raft

Re-arranging the equation and substituting the values, we find: the final velocity of the canoe:

$v_1 = \frac{m_1 u_1 + m_2 u_2-m_2 v_2}{m_1}=\frac{(16.0)(-12.5)+(14.0)(16.0)-(14.0)(-14.4)}{16.0}=+14.1 m/s$

So, the canoe is moving at 14.1 m/s to the right after the collision.

Learn more about momentum:

brainly.com/question/7973509

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5 0

3 years ago

According to collision theory, which condition(s) must be met in order for molecules to react?

kondaur [170]

Answer:

B because molecules occurs by the reaction

5 0

3 years ago