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dybincka [34]
3 years ago
7

10. How much total work do you do when you lift a 50 kg microwave 1.0 m off the ground and then push it 1.0 m

Physics
1 answer:
Mariulka [41]3 years ago
6 0

Work formula:

W = Fd\cos(\theta)

F = 50N, d = 1.0 m

When you lift something straight up, the angle of the force is 90º

cos(90º) is 0, so there's no work done when you lift the microwave off the ground

W = (50N)(1.0)(0) = 0

F = 50N, d = 1.0 m

When you push the microwave, the angle is 0º and cos(0º) is 1. So there is work done here:

W = (50 N)(1.0m)(1)

W = 50

total work = 50 joules

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How much does the gravitational force of attraction change between two asteroids if the two asteroids drift three times closer t
Katen [24]

Answer:

Increase 9 times

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 \times 10^{-11} m^3/kgs^2 is the gravitational constant. M_1, M_2 is the masses of the 2 objects. and R is the distance between them.

Since the force is inversely proportional to the distance squared, if it is reduced by 3 times, the gravitational force between them would increase by 3^2 = 9 times

6 0
3 years ago
The position of a ball rolling in a straight line is given by x=2−3.6t+4.4t2, where x is in meters and t in seconds.
kobusy [5.1K]
I'm assuming the question is time it will take for ball to reach ground, if it is then set equation to zero then use the quadratic formula, the possible t value is your answer then
8 0
3 years ago
A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0
3 years ago
The velocity of a body is increases from 10 m/s ti 15 m/s in 5seconds calculate its acceleration​
Veseljchak [2.6K]

Answer:

acceleration = v-u/ t

= 15-10/5

= 5/5

= 1 m/s2

Explanation:

hope this helped you.

7 0
3 years ago
Read 2 more answers
A sample of an unknown substance has a mass of 0. 158 kg. If 2,510. 0 J of heat is required to heat the substance from 32. 0°C
guapka [62]

The specific heat of the unknown substance with a mass of 0.158kg is 0.5478 J/g°C

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

The specific heat capacity of a substance can be calculated using the following formula:

Q = m × c × ∆T

Where;

  • Q = quantity of heat absorbed (J)
  • c = specific heat capacity (4.18 J/g°C)
  • m = mass of substance
  • ∆T = change in temperature (°C)

According to this question, 2,510.0 J of heat is required to heat the 0.158kg substance from 32.0°C to 61.0°C. The specific heat capacity can be calculated:

2510 = 158 × c × (61°C - 32°C)

2510 = 4582c

c = 2510 ÷ 4582

c = 0.5478 J/g°C

Therefore, the specific heat capacity of the unknown substance that has a mass of 0.158 kg is 0.5478 J/g°C.

Learn more about specific heat capacity at: brainly.com/question/2530523

4 0
3 years ago
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