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Elan Coil [88]
3 years ago
7

An object has a mass of 8.00kg. What is the gravitational force on the object by the earth

Physics
1 answer:
Firdavs [7]3 years ago
6 0
78.4 F because you do 8.00 muliplyed by 9.8
You might be interested in
How much time passes when an object travels at a constant speed of 17 m/s over a distance of 323 m? Identify Variables, write fo
k0ka [10]

Answer: 19 s

Explanation:

The relationship between speed, distance  and time is given by:

velocity= distance/time

time= distance/velocity= 323 m/17 m/s= 19 s

8 0
3 years ago
A cylindrical metal specimen having an original diameter of 11.77 mm and gauge length of 46.1 mm is pulled in tension until frac
Marta_Voda [28]

Answer:

% reduction in area = 54.26 %

percentage elongation = 43.16 %

Explanation:

a) percentage reduction in area = \frac{(A_1 - A_2)}{A_1 } * 100

A_1  =\pi r^2 = \pi * 5.88^2=108.8 mm2

A_2=\pi * 3.98^2= 49.97 mm2

% reduction in area =\frac{(108.8 - 49.97)}{108.8} * 100 = 54.26 %

b)percentage elongation = \frac{(66 - 46.1)}{46.1} *100 = 43.16 %

5 0
3 years ago
2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her he
Ratling [72]

Answer:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, v_s = d/t₂

∴ v_s = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s

The minimum average speed the opponent must move so that he is in position to hit the ball, v_s ≈ 5.79 m/s.

5 0
3 years ago
Need help girliess.
steposvetlana [31]
If I remember correctly, it is the 3rd answer choice.
8 0
3 years ago
A 0.125 kg hockey puck moving at 20.0 m/s is caught and held by an 85.0 kg goalkeeper at rest. After catching the puck, with wha
jenyasd209 [6]

Answer:

a

Explanation:cuz i saiad so

4 0
3 years ago
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