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mrs_skeptik [129]

4 years ago

6

HELP PLEASE! 98 POINTS FOR ANSWER

1 answer:

sasho [114]4 years ago

3 0

Well! see the car experiences 300 N force to the right from the engine.

It also experiences 150N force to left due to friction and air resistance

Now, we know it is the net force which will decide the car will move in which direction

We have net force=300-150=150N

which is towards right

So that means car will accelerate in right direction.

Hope it helps!!

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You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.How muc

ahrayia [7]

Work is equal to the force applied times the displacement. Since you pull the wagon at constant speed this means that there is no acceleration on the wagon as it does not change speed. F=ma. Since a=0, F=0. Therefore no work has been done in this situation

3 0

3 years ago

1. A 0.40 kg ball is launched at a speed of 16 m/s from a 22 m cliff.

Masja [62]

Answer:

51.2 J, 86.2 J, 137.4 J

Explanation:

The kinetic energy of the ball is given by:

$K=\frac{1}{2}mv^2$

where

m = 0.40 kg is its mass

v = 16 m/s is its speed

Substituting,

$K=\frac{1}{2}(0.40)(16)^2=51.2 J$

The potential energy of the ball is given by

$U=mgh$

where

m = 0.40 kg

$g=9.8 m/s^2$ is the acceleration of gravity

h = 22 m is the heigth of the cliff

Substituting,

$U=(0.40)(9.8)(22)=86.2 J$

Finally, the total mechanical energy is the sum of the kinetic energy and the potential energy:

$E=K+U=51.2 + 86.2=137.4 J$

4 0

3 years ago

Define malleability?<br>

ad-work [718]

Answer:

Explanation:

the quality or state of being malleable: such as. a : capability of being shaped or extended by hammering, forging, etc. the malleability of tin.

5 0

3 years ago

Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of

Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = $\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}$

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = $\sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}$

(|r₂₁|)² = 148.25

$F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }$

= 0.050938(0.19107i + 0.54933j) N

= (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0

3 years ago

What is the thermal energy of an object?

Vedmedyk [2.9K]

The total kinetic and potential energies

3 0

3 years ago