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Hatshy [7]
3 years ago
10

5. The force of ________ is the force at which the earth attracts another object towards itself.

Physics
2 answers:
iragen [17]3 years ago
7 0
(answer )---> C-gravity
dangina [55]3 years ago
7 0
Gravity is the source of the equal forces with which the Earth and another object attract each other.
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Sound waves have a frequency of 1000 hz what is their period?​
mr_godi [17]

Answer:

Solve it

Explanation:

The formula for time is: T (period) = 1 / f (frequency). λ = c / f = wave speed c (m/s) / frequency f (Hz). The unit hertz (Hz) was once called cps = cycles per second.

7 0
3 years ago
In an experiment, a solid, uniform sphere is at rest on a horizontal surface. A net force is applied tangentially to the edge of
Thepotemich [5.8K]

Answer:

Option A and D

Radius of the sphere, and,

Average net force exerted on the sphere from 0 to 3s

Explanation:

The product of the average net force and the radius will give the torque

T = F x r

The product of torque and time will give the change in angular momentum

T x t = I2w2 - I1w1

Where F is the average force,

r is the radius,

T is the torque,

t is the time (0 to 3sec)

I = moment of inertia of the system

w is the angular speed.

4 0
3 years ago
WHICH SHOULD NOT BE PART OF SCIENTIFIC INQUIRY
umka21 [38]

Answer:

Not collecting the info

Explanation:

5 0
3 years ago
What force always resists or apposes acceleration?
Mariana [72]
Air always resists or apposes acceleration
7 0
3 years ago
A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 4.0 m/s, skat
Cerrena [4.2K]

Answer:

a)t=59.817652833125294s \approx 59.8s

b)D_1=894.01m

Explanation:

From the question we are told that

Speed of opposing player V_2=4.0m/s

First player chase his opponent aftert=2.80t

Acceleration of  first player a=0.14 m/s2

Let time of catch be  t_c

a)Generally the Equation for distance covered is mathematically given as follows

Distance to catch First opponent

D_1=\frac{1}{2}(0.14)t^2

D_1=0.07t^2

Distance to covered Second opponent

D_2=(2.8+t)*4

Generally when first opponent catch the second opponent it is represented mathematically as

D_2=D_1

0.07t^2=(2.8+t)*4

0.07t^2=11.2+4t

0.07t^2-4t-11.2

t=59.817652833125294s \approx 59.8s

b)Generally the the total time traveled by the first opponent is mathematically given as

D_1=\frac{t^2}{4}

D_1=\frac{59.8^2}{4}

D_1=894.01m

3 0
3 years ago
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