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kolbaska11 [484]

3 years ago

11

If a sled has at the top of 10 m hill had 1000 J of potential energy what would happen to the PE if the sled were to moved to a

5 m hill?

1 answer:

nataly862011 [7]3 years ago

7 0

Answer:

500J

Explanation:

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there are a wats group where you can get your answer by trusted experts.

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A hiker walks 20.51 m at 33.16 degrees. What is the Y component of his displacement?

Serhud [2]

Answer:

<em>The y component of his displacement is 11.22 meters</em>

Explanation:

<u>Components of the displacement</u>

The displacement is a vector because it has a magnitude and a direction. Let's suppose a displacement has a magnitude r and a direction θ, measured with respect to the positive x-direction. The horizontal component of the displacement is calculated by:

$x=r\cos\theta$

The vertical component is calculated by:

$y=r\sin\theta$

The hiker has a displacement with magnitude r = 20.51 m at an angle of 33.16 degrees. Substituting in the above equation:

$y=20.51\sin(33.16^\circ)$

$y=11.22\ m$

The y component of his displacement is 11.22 meters

7 0

3 years ago

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

Oksana_A [137]

Answer:

a.Attractive

$2.31531\times 10^{-16}\ N$

Explanation:

When it comes to charges, the charges which are alike repel each other and the charges which are different will attract each other.

Here, there is a proton and electron which are different particles hence, they will attract each other.

$q_1=q_2$ = Charge of electron and proton = $1.6\times 10^{-19}\ C$

r = Distance between them = 997 nm

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(997\times 10^{-9})^2}\\\Rightarrow F=2.31531\times 10^{-16}\ N$

The force of attraction between the particles will be $2.31531\times 10^{-16}\ N$

8 0

3 years ago

A roller coaster car rapidly picks up speed as it rolls down a slope as it starts down the slope its speed is 4m/s but 3 seconds

Gwar [14]

Answer:

The acceleration is 6 [m/s^2]

Explanation:

We can find the acceleration of the roller coaster using the kinematic equation for uniformly accelerated motion.

$v_{f} =v_{i} + a*t\\where:\\v_{f} = final velocity = 22 [m/s]\\v_{i} = initial velocity = 4 [m/s]\\t = time = 3 [s]\\$

Now replacing the values we have:

$a=\frac{v_{f} - v_{i} }{t} \\a=\frac{22 - 4 }{3}\\a = 6 [m/s^{2} ]$

3 0

3 years ago

Look at the resistor illustrated in the figure above. Based on your knowledge of the resistor color code, what are the resistanc

vichka [17]

A because of the resistors are four in this options first option is multiplied by 4

6 0

3 years ago

Read 2 more answers

A Cessna aircraft has a lift off speed of 147 km/h. What minimum constant acceleration does this require if the aircraft is to b

netineya [11]

Answer:

The acceleration is $a =51945 \ km/h^2$

Explanation:

From the question we are told that

The lift up speed is $v = 147 \ km/h$

The distance covered for the take off run is $s = 208 m = 0.208 \ km$

Generally from kinematic equation we have that

$v^2 = u^2 + 2as$

Here u is the initial speed of the aircraft with value 0 m/ s give that the aircraft started from rest

So

$147^2 = 0^2 + 2* a* 0.208$

=> $a =51945 \ km/h^2$

6 0

3 years ago