All categories

2 years ago

How high is a 0.5 kg squirrel sitting if it has 36 j of energy

Physics

2 answers:

iVinArrow [24]2 years ago

7 0

The squirrel is sitting at a height of 7.35 m.

The energy possessed by a body by virtue of its position is called potential energy.

If the squirrel has a mass m and if it sits at a height h then, the potential energy E of the squirrel is given by,

$E=mgh$

here, g is the acceleration due to gravity.

Given:

$m=0.5 kg;E=36 J$

Take the value of acceleration due to gravity $g=9.8 m/s^2$

Rewrite the equation for energy in terms of h.

$h=\frac{E}{mg}$

Substitute the given values and calculate the value of h.

$h=\frac{E}{mg} \\ =\frac{36 J}{(0.5 kg)(9.8 m/s^2)} \\ =7.35 m$

The squirrel is at a height of 7.35 m

jekas [21]2 years ago

7 0

Assuming potential energy. PE=mgh
7.35 meters above ground

You might be interested in

Vector A⃗ has magnitude 8.00 m and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axi

WINSTONCH [101]

Answer:

293.7 degrees

Explanation:

A = - 8 sin (37) i + 8 cos (37) j

A + B = -12 j

B = a i+ b j , where and a and b are constants to be found

A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

- 12 j = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

Comparing coefficients of i and j:

a = 8 sin (37) = 4.81452 m

b = -12 - 8cos(37) = -18.38908

Hence,

B = 4.81452 i - 18.38908 j ..... 4 th quadrant

Hence,

cos ( Q ) = 4.81452 / 12

Q = 66.346 degrees

360 - Q = 293.65 degrees from + x-axis in CCW direction

5 0

2 years ago

A man holding 7N weight moves 7m horizontal and 5m vertical , find the work done

SashulF [63]

Answer:

35 J

Explanation:

The man is holding the box: this means that he is applying a force vertically upward, to balance the weight of the box (which pushes downward).

Therefore, we can ignore the horizontal displacement of the man, because the force applied (vertically upward) is perpendicular to that displacement (horizontal), so the work done for that is zero.

So, only the vertical motion contributes to the work. The work done by the man is equal to the gain in gravitational potential energy of the box, so:

$W=(mg)\Delta h$

where

$mg=7 N$ is the weight of the box

$\Delta h=5 m$ is the vertical displacement

Substituting, we find

$W=(7N)(5 m)=35 J$

8 0

2 years ago

Which statement would be a valid argument in favor of using nuclear power? There are few negative impacts from mining the fuel.

fgiga [73]

Okay so, lets use the process of elimination here.

A)There are few negative impacts from mining the fuel.
B)Reactors are safe from natural disasters.
C)There are little to no waste products from fission.
D)Nuclear power does not contribute greenhouse gases.

First off, we know B cannot be correct, seeing as how reactors are fragile and are damaged easily by Japan's earthquakes. So we can eliminate B from the choices. We then can eliminate C, since fission creates high levels of nuclear waste, so that leaves us with just A, and D. We can then eliminate A since uranium is radioactive, there is always a chance for negative effects.

So, the correct answer is D

6 0

2 years ago

Read 2 more answers

A siphon pumps water from a large reservoir to a lower tank that is initially empty. The tank also has a rounded orifice 20 ft b

trasher [3.6K]

Answer:

height of the water rise in tank is 10ft

Explanation:

Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)

$\frac{P_1}{pg} + \frac{V^2_1}{2g} + z_1= \frac{P_2}{pg} + \frac{V_2^2}{2g} +z_2$

$\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}$ -------(1)

substitute $P_a_t_m for P_1, (P_a_t_m +pgh) for P_2$

0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)

$\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}$

$\frac{P_1}{pg} + \frac{0^2_1}{2g} +( 20)= \frac{(P_a_t_m+pgh)}{pg} +\frac{V^2_2}{2\times32.2} \\\\V_2 = \sqrt{64.4(20-h)}$

Applying bernoulli's equation between tank surface (3) and orifice exit (4)

$\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4$

substitute

$P_a_t_m for P_3, P_a_t_m for P_4$

0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g

$\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}$

At equillibrium Fow rate at point 2 is equal to flow rate at point 4

Q₂ = Q₄

A₂V₂ = A₃V₃

The diameter of the orifice and the siphon are equal , hence there area should be the same

substitute A₂ for A₃

$\sqrt{64.4(20-h)}$ for V₂

$\sqrt{64.4h}$ for V₄

A₂V₂ = A₃V₃

$A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft$

Therefore ,height of the water rise in tank is 10ft

3 0

2 years ago

You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.

Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height y₁ = 800m

Angle θ = 25°

cos 25 = x / r

sin 25 = z / r

x₁ = r cos 20

z₁ = r sin 25

x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height y₂ = 1100 m

Angle θ = 20°

x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

d² = (18126-16314)² + (1100-800)² + (8452-7607)²

d² = 3,283 106 +9 104 + 7,140 105

d² = (328.3 + 9 + 71.40) 10⁴

d = √(408.7 10⁴)

d = 20,216 10² m

d = 2021.6 km

7 0

2 years ago