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4 years ago

What is the relationship between mass and gravitational force between any two objects?

1 answer:

7 0

If two objects' mass increases, gravitational force pulls them down to the center more strongly because the greater the mass, the stronger the attraction is from the gravitational force from any two objects. An increase in then Earth's mass and increase in a man's mass would make gravitational force pull more strongly on the person because the Earth's gravitational force multiplied from the mass change and the man's greater mass pulls them down more quickly.

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If a flea can jump straight up to a height of 0.410 m , what is its initial speed as it leaves the ground?

aivan3 [116]

Initial velocity = $v_0$

acceleration in the downward direction = -9.8 $\frac {m}{s^2}$

Final velocity at the highest point = 0

Maximum height reached = 0.410 m

Now, Using third equation of motion:

$\(v^2 = {v_0}^{2} + 2aH$

$\(0^2 = {v_0}^{2} - 2 \times 9.8 \times 0.410$

${v_0}^{2} = 2 \times 9.8 \times 0.410$

$v_0 = 2.834 \frac {m}{s}$

Speed with which the flea jumps = $2.834 \frac {m}{s}$

4 0

4 years ago

One wire carries a current of I1= 2 Amps, and the other carries a parallel current (same direction, wires are side by side) of I

Yuki888 [10]

Answer: $F=3\times 10^{-5} N$

Explanation:

Given

$I_1=2 Amps$

$I_2=0.75 Amps$

Length of each wires $\left ( L\right )=5 m$

Distance between wires $\left ( r\right )=5 cm$

Force per unit length = $\frac{\mu_0I_1I_2}{2\pi r}$

$F'=\frac{2\times 10^{-7}\times 2\times 0.75}{2\pi \times 0.05}$

$F'=6\times 10^{-6}$

Force for $L=5 m$

$F=3\times 10^{-5} N$

6 0

3 years ago

Will mark brainliest if answer

Anvisha [2.4K]

I won't touch the object because there is a warning sign and that means you stay out if it because it is a danger sign that High Resistance Material that means it is dangerous.

3 0

2 years ago

A body moves in a straight line. At time, it's acceleration is given by a = 12t + 1. When t =0, the velocity of the body v is 2

madreJ [45]

Answer:

v = 6t² + t + 2, s = 2t³ + ½ t² + 2t

59 m/s, 64.5 m

Explanation:

a = 12t + 1

v = ∫ a dt

v = 6t² + t + C

At t = 0, v = 2.

2 = 6(0)² + (0) + C

2 = C

Therefore, v = 6t² + t + 2.

s = ∫ v dt

s = 2t³ + ½ t² + 2t + C

At t = 0, s = 0.

0 = 2(0)³ + ½ (0)² + 2(0) + C

0 = C

Therefore, s = 2t³ + ½ t² + 2t.

At t = 3:

v = 6(3)² + (3) + 2 = 59

s = 2(3)³ + ½ (3)² + 2(3) = 64.5

5 0

3 years ago

You need to calculate the volume of berm that has a starting cross-sectional area of 118 SF, and an ending cross-sectional area

LekaFEV [45]

6 cubic feet I’m pretty sure that’s the answer

8 0

3 years ago