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3 years ago

1. Define weight

1 answer:

8 0

1. Weight is the gravitational pull with which the earth attracts the body towards the center of the earth. The S.I unit is Newton (N)

2. The weight of an object is related to its mass with the below equation.

W = mg

Where W = weight, m = mass and g = acceleration due to gravity

3. The mass m = 50 kg and g = 9.8 m/ $s^{2}$

Substitute the parameters in the equation above.

W = 50 x 9.8

W = 490 N

4. An object is accelerating when it speeds up. If the object slows down, it means it is decelerating. The correct answer is option A

5. Newton's second law of motion is:

F = ma

Where F = force applied, m = mass and a = acceleration

Therefore, Newton's second law of motion relates an object's acceleration to its net force acting on it. The correct answer is option C

6. According to Newton's second law of motion which relates an object's acceleration to its mass, doubling the net force acting on an object a doubles its acceleration. Because mass is always constant.

7. Since the weight of an object is related to its mass with the equation.

W = mg, If the mass of an object doubles, its weight will also doubles. Option A is the correct answer.

8. If you know the mass of an object, you can calculate its weight with the formula F = mX when X = 9.8m/ $s^{2}$

9. Force is expressed in unit as Newton (N)

10. The parameters given are :

mass m = 20kg

Force F = 40N

To calculate acceleration, use the formula F = ma

Substitute all the parameters into the equation.

40 = 20a

a = 40/20

a = 2m/ $s^{2}$

The correct option is D

Learn more here : brainly.com/question/18835375

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Lina20 [59]

Answer:

Acoustic microstreaming

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3 years ago

An electron moves with velocity v⃗ =(5.8i−6.7j)×104m/s in a magnetic field B⃗ =(−0.81i+0.60j)T.

Minchanka [31]

Answer:

Fₓ = 0, $F_{y}$ = 0 and $F_{z}$ <em> = - 3.115 10⁻¹⁵ N</em>

Explanation:

The magnetic force given by the expression

F = q v xB

the bold are vectors, the easiest analytical way to determine this force in solving the determinant

$F = q \left[\begin{array}{ccc}i&j&k\\5.8&-6.7&0\\-0.81&0.6&0\end{array}\right] 10^{4}$

F = 1.6 10⁻¹⁵ [ i( 0-0) + j (0-0) + k^( 5.8 0.60 - 0.81 67) ]

F =i^0 + j^0 - k^ 3.115 10⁻¹⁵ N

Fₓ = 0

$F_{y}$ = 0

$F_{z}$ <em> = - 3.115 10⁻¹⁵ N</em>

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3 years ago

What is short circuit

Marizza181 [45]

Answer:

A short circuit is an electrical circuit that allows

Explanation:

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3 years ago

Read 2 more answers

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

uranmaximum [27]

Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

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3 years ago

Question 6

Effectus [21]

Answer:

4.9 x 10-23 N

Explanation:

I took the test

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3 years ago