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den301095 [7]
2 years ago
7

An electric motor has a rating of 4.0 x 10^2 watts. How much time will it take for this motor to lift a 50.-kilogram mass a vert

ical distance of 8.0 meters?
Physics
1 answer:
LenKa [72]2 years ago
4 0
ΔT = ?

P = 50 kg

ΔS = 8.0 m

g = 9.8 m/s²

Pot = 4,0x10² W

Find the time :

ΔT = P * g *  ΔS / Pot

ΔT = 50 * 9.8 * 8.0 / 4.0x10²

ΔT = 3920 / 4x10²

ΔT = 9.8 S

hope this helps!


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The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from
Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

7 0
1 year ago
A teacher sets up a stand carrying a convex lens of focal length 15 cm at 20.5 cm mark on the optical bench. She asks the studen
Brums [2.3K]
We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I  attached we can see that:
\frac{h_i}{h_0}=\frac{d_i}{d_0}
Two triangles that I marked are similar and from this we get:
\frac{h_i}{h_0}=\frac{d_i-f}{f}
The image and the object must have the same height so we get:
\frac{h_i}{h_0}=\frac{d_i-f}{f};h_i=h_0\\
1=\frac{d_i-f}{f}\\
d_i=2f
This tells how far the screen should be from the lens. 
The position of the screen on the optical bench is:
S=20.5cm+2f=20.5+2\cdot 15cm=50.5cm

8 0
1 year ago
1. Each year at a college, there is a tradition of having a hoop rolling competition. Alex rolls his 0.350 kg hoop down the cour
grigory [225]

Question 1:

Answer:

The moment of inertia of Alex's rolling hoop is 0.197 kg \cdot cm^2

Explanation:

<u>Given</u>:

Mass of the hoop = 0.350 g

Radius of the hoop = 75.0 cm

<u>To Find:</u>

The moment of inertia of Alex's rolling hoop = ?

<u>Solution</u><u>:</u>

The moment of inertia  = mr^2

where

m is the mass

r is the radius

Converting cm to m, we get

75.0 cm = 0.75 m

Now substituting the values,

=> moment of inertia  = (0.350)(0.75)^2

=> moment of inertia  = (0.350)(0.5625)

=> moment of inertia  = (0.197)

Question 2:

Answer:

The combined angular momentum of the masses is 1.76 kg m^2 s^{-1}

If she pulls her arms in to 0.12 m, her new linear speed  is  18.33 m/s^2

Explanation:

Given:

Mass  = 2.0 kg

Radius = 0.8 m

Velocity =  1.2 m/s

a.The combined angular momentum of the masses:

L = r \cdot m \cdot v_1

Substituting the values,

L = 0.8 \cdot 2.0 \cdot 1.1

L= 1.76 kg m^2 s^{-1}

b. If she pulls her arms in to 0.12 m, what is her new linear speed

0.12 \cdot 0.8 \cdot v_2 = 1.76

0.096 cdot v_2 = 1.76

v_2 = \frac{1.76}{0.096}

v_2 = 18.33 m/s^2

6 0
2 years ago
Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth.
marishachu [46]

Answer:

h=\frac{1}{2}\frac{v^2}{g}

Explanation:

Let's assume that an object is launched straight upward in a gravitational field. Its initial kinetic energy is given by

K=\frac{1}{2}mv^2 (1)

where m is the mass and v is the initial speed.

As the object goes higher, its kinetic energy decreases and it is converted into gravitational potential energy, since the total mechanical energy (sum of kinetic and potential energy) must remain constant:

E=K+U=const.

At the highest point of the trajectory, the speed of the object is zero (v=0), so the kinetic energy is also zero (K=0), which means that all the kinetic energy has been converted into potential energy:

U=mgh (2)

where g is the gravitational acceleration and h is the maximum height of the object.

Due to conservation of energy, we can write that (1) and (2) are equal, so:

\frac{1}{2}mv^2=mgh

from which we can derive an expression for the maximum height reached by the object

h=\frac{1}{2}\frac{v^2}{g}

5 0
2 years ago
As a youngster, you drive a nail in the trunk of a young tree that is 3 meters tall. The nail is about 1.5 meters from the groun
Lera25 [3.4K]

Answer:

15m

Explanation:

Hello! first to solve this problem we must find that so much that the tree grew in the 15 years this is achieved by dividing the height of the tree before and after

\frac{30m}{3m} =10 times

the tree grew 10 times its initial length in 15 years, so to find how tall the nail is, we multiply this factor by 1.5m

X=(1.5m)(10)=15m

the nail is 15 meters above ground level

8 0
2 years ago
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