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ivann1987 [24]
3 years ago
12

What it the direction of the weight or force of gravity?

Physics
1 answer:
AveGali [126]3 years ago
3 0
Toward the center of the earth
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Suppose you lift a 20 kg box by a height of 1.0 m.
9966 [12]
W=mgh W=(20)(9.8)(1) w=196J
6 0
2 years ago
Read 2 more answers
Find the work w1 done on the block by the force of magnitude f1 = 95.0 n as the block moves from xi = -5.00 cm to xf = 4.00 cm .
Vlad [161]
<h3><u>Answer;</u></h3>

= 8.55 Joules

<h3><u>Explanation;</u></h3>

Work done is the product of force and the distance moved by an object.

Work done = Force × distance

Force = 95 Newtons

Distance = X2 -X1

               = 4 - (-5)

               = 9 cm

Thus;

work done = 95 × 9/100

                  <u>= 8.55 Joules </u>

5 0
3 years ago
FREE BRAINIEST IF YOU ANSWER THIS
Inessa [10]

The watt is a rate, similar to something like speed (miles per hour) and other time-interval related measurements.

Specifically, watt means Joules per Second. We are given that the electrical engine has 400 watts, meaning it can make 400 joules per second. If we need 300 kJ, or 3000 Joules, then we can write an equation to solve the time it would take to reach this amount of joules:

w * t = E

w: Watts

t: Time

E: Energy required

(Watts times time is equal to the energy required)

<u>Input our values:</u>

400 * t = 3000

(We need to write 3000 joules instead of 300 kilojoules, since Watts is in joules per second. It's important to make sure your units are consistent in your equations)

<u>Divide both sides by 400 to isolate t:</u>

<u />\frac{400t}{400} = \frac{3000}{400}

t = 7.5 (s)

<u>It will take 7.5 seconds for the 400 W engine to produce 300 kJ of work.</u>

<u></u>

If you have any questions on how I got to the answer, just ask!

- breezyツ

6 0
2 years ago
Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach
Svetlanka [38]

Answer:

\Delta V    = 1.8 \times 10^7 V

Explanation:

GIVEN

diameter = 15 fm  =15 \times 10^{-15}m

we use here energy conservation

K_{i}+U_{i} =K_{f}+U_{f}

there will be some initial kinetic  energy but after collision kinetic energy will zero

K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}

on solving these equations we get kinetic energy initial

KE_{i} = 5.65\times 10 ^{-12} \times \frac  {1 eV}{1.6 \times 10^{-19}}

KE_{i} = 35.33 J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the {238}_U nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

\Delta V = \frac{35.33}{2}  MV\\\Delta V    = 1.8 \times 10^7 V

7 0
3 years ago
A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th
son4ous [18]

Answer:

Part a)

v = 7.57 km/h

Part b)

\theta = 67.5 degreeNorth of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

d_1 = 40 km

now it turns towards 50 degree East of North

so its distance is given as

d_2 = 20 km(sin50 \hat i + cos50\hat j)

d_2 = 15.3 \hat i + 12.8 \hat j

then finally it moves towards west for 50 min

d_3 = -50 \hat i

Now the total displacement of the train is given as

d = d_1 + d_2 + d_3

d = (40 + 15.3 - 50)\hat i + 12.8 \hat j

d = 5.3\hat i + 12.8 \hat j

now total time duration of the motion is given as

T = 40 min + 20 min + 50 min

T = 1.83 h

now average velocity is given as

v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}

v_{avg} = 2.89\hat i + 6.99\hat j

Part a)

magnitude of the average velocity is given as

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{2.89^2 + 6.99^2}

v = 7.57 km/h

Part b)

Direction of the velocity is given as

tan\theta = \frac{v_y}{v_x}

tan\theta = \frac{6.99}{2.89}

\theta = 67.5 degreeNorth of East

6 0
3 years ago
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