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4 years ago

Can someone please help me? Will give brainliest.

Physics

2 answers:

lora16 [44]4 years ago

5 0

Needs are less than resources, and the population increases.

Soloha48 [4]4 years ago

3 0

Needs are less than resources

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marshall27 [118]

H<span>igh voltage demonstrations!</span>

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3 years ago

Read 2 more answers

The car's motor transfers

Elanso [62]

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chemical potential

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3 years ago

An 8.75 kg point mass and a 14.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be

marysya [2.9K]

Answer

given,

mass of the = m₁ = 8.75 Kg

another mass of the object = m₂ = 14 Kg

distance between them = 50 cm

R₁ = 17 cm

R₂ = 50 -17 = 33 cm

a) Force applied due to the Mass 8.75 in +ve x- direction

$F_1 = \dfrac{GM_1 m}{R_1^2}$

$F_1 = \dfrac{6.67\times 10^{-11} \times 8.75\ m}{0.17^2}$

$F_1 = 2.019\times 10^{8}\ m$

Force applied due to mass 14 Kg in -ve x-direction

$F_2 = \dfrac{GM_2 m}{R_2^2}$

$F_2 = \dfrac{6.67\times 10^{-11} \times 14\ m}{0.33^2}$

$F_2 = 0.857\times 10^{8}\ m$

net force

F = F₁ + F₂

$F = 2.019\times 10^{8}\ m - 0.857\times 10^{8}\ m$

$F = 1.162\times 10^{8}\ m$

Using newton second law

$a = \dfrac{F}{m}$

$a = \dfrac{ 1.162\times 10^{8}\ m}{m}$

$a =1.162\times 10^{8} \ m/s^2$

b) As the acceleration of mass comes out to be +ve hence, the direction will be toward the mass of 8.75 Kg

6 0

4 years ago

A ball is thrown upward at time t=0 from the ground with an initial velocity of 60 m/s (~ 134 mph). Assume that g = 10 m/s2.At w

Shtirlitz [24]

Answer:

6 second

Explanation:

initial velocity of ball, u = 60 m/s

g = 10 m/s^2

Let the ball takes time t to reach at the maximum height

We know that at maximum height, the velocity of ball is zero.

v = 0 m/s

Use first equation of motion

v = u + gt

0 = 60 - 10 x t

t = 6 second

Thus, the ball takes 6 second to reach to maximum height.

7 0

3 years ago

A speedboat travels at an average of 20ms-1 calculate the distance it will cover in 25mins

vladimir1956 [14]

Answer:

30000m

Explanation:

25mins = 25 × 60

=1500secs

Let total distance=D

Average speed= total distance÷total time

20m/s= D÷1500secs

Cross multiplying

20×1500= D

=30000m

5 0

1 year ago