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Katena32 [7]
3 years ago
12

PLEASE ASAP ILL GIVE BRAINLIEST.

Physics
2 answers:
Sergeu [11.5K]3 years ago
6 0

Answer:

law of gravity

Explanation:

cause the ball was still moving

Debora [2.8K]3 years ago
5 0

Answer: the answer is law of gravity

Explanation:

You might be interested in
How much net force is required to accelerate a 2000 kg car at 3.00 m/s^2
andrezito [222]

The net force required to accelerate a car is 6000 N.

Force is defined as the product of the mass and acceleration of the body. Force is used to changing the velocity that is to accelerate an object or a body of a particular mass. The unit of Force is Newton or kg m/s^2.

The formula used to calculate the net force is :

F = ma

where, F = Force

m = mass = 2000 kg

a = acceleration = 3.00 m/s^2

∴ F = 2000*3

F = 6000 N

Thus, to accelerate the car at 3.00 m/s^2 of mass 2000 kg net force required is 6000 N.

To learn more about force,

brainly.com/question/1046166

6 0
1 year ago
What factors affect water quality?
JulsSmile [24]

Answer:

Explained below:

Explanation:

Water quality is determined in terms of the physical, chemical and biological content of water. The quality of water of lakes and rivers variates with the seasons and also geographic areas, still in the absence of pollution. There are  so many factors  which affect water quality are as follows :

Pesticides

Temperature

Runoff

Sedimentation

Erosion

pH

Dissolved oxygen

Pesticides

Litter and rubbish

Decayed organic materials

Toxic and hazardous substances

Oils, grease, and other chemicals

3 0
3 years ago
A ticker timer makes 50 dots per second. When a body is pulled by a tap through the timer, the distance between the third and fo
fenix001 [56]

Answer:

The acceleration is 1 cm/s^2.

Explanation:

The acceleration is defined as the rate of change of velocity.

Here, initial velocity, u = 3/1 = 3 cm/s

final velocity, v = 4/1 = 4 cm/s

time, t = 1 s

Let the acceleration is a.  

Use first equation of motion

v =  u + at

4 = 3 + 1 x a

a = 1 cm/s^2

8 0
2 years ago
The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discover
givi [52]

Answer:

Acceleration due to gravity, a=4.61\times 10^{-14}\ m/s^2

Explanation:

It is given that,

Mass of Pluto, m=1.4\times 10^{22}\ kg

Distance between Neptune and Pluto, r=4.5\times 10^{12}\ m

The force of gravity is balanced by the gravitational force between Neptune and Pluto. It is given by :

a=\dfrac{Gm}{r^2}

a=\dfrac{6.67\times 10^{-11}\times 1.4\times 10^{22}}{(4.5\times 10^{12})^2}

a=4.61\times 10^{-14}\ m/s^2

So, the acceleration due to gravity at Neptune due to Pluto is 4.61\times 10^{-14}\ m/s^2. Hence, this is the required solution.  

4 0
3 years ago
Block B (of mass m) is initially at rest. Block A (of mass 3m) travels toward B with an initial speed v0 (vee nought) and collid
NeTakaya

Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

The mass of block B = m

The mass of block A = 3·m

The initial velocity of block B, v₂ = 0 m/s

The initial velocity of block A, v₁ = v₀

The amount of friction between the blocks and the surface = Negligible friction

By the Law of conservation of linear momentum, we have;

Total initial momentum = Total final momentum

3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃

Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

∴ 3·m × v₀ =  4·m·v₃

\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2

The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

The potential energy, P.E. = m·g·h

Where;

m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

m \cdot g \cdot h =  \dfrac{9}{8} \cdot m\cdot v_0^2

h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g }  = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g }  =  \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot  v_0^2 \approx 0.1147959   \cdot  v_0^2

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

7 0
2 years ago
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