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3 years ago

11

A 30 N rightward force is exerted on a 7 kg object. There is a 3 N force of friction acting on the object. What is the Weight of

the Object?

1 answer:

IgorLugansk [536]3 years ago

6 0

Answer:

W = 68.6 N

Explanation:

Given that,

Force acting on a right side = 30 N

Mass of the object, m = 7 kg

The force of friction acting on the car = 3 N

We need to find the weight of the object.

The weight of an object is given by :

W = mg

Substitute all the values,

W = 7 × 9.8

W = 68.6 N

So, the weight of the object is 68.6 N.

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Korolek [52]

Answer:

7.5 m/s2

Explanation:

F = ma

300 = 40 kg (a)

divide both sides by 40 to single out the variable a (acceleration)

300/40 = 7.5 m/s2

6 0

3 years ago

What is Physics? How is Physics used in your everyday life? Give at least two examples and explain your answer. You need at leas

NNADVOKAT [17]

Answer:

Physics is a brand of science that deals with nature, matter, and energy. Physics is used in the everyday world because we have simple objects such as levers that use physics. we also have Newton's law of transportation that manipulate everyday objects making them move, so without physics many objects wouldn't move.

3 0

3 years ago

At t = 0, Ball 1 is dropped from the top of a 22 m-high building. At the same instant Ball 2 is thrown straight up from the base

Crazy boy [7]

Answer:

<em>The two balls pass each other at a height of 5.53 m</em>

<em>vf1=17.97 m/s</em>

<em>vf2=-5.96 m/s</em>

Explanation:

<u>Vertical Motion</u>

An object thrown from the ground at speed vo, is at a height y given by:

$y=vo.t-g.t^2/2$

Where t is the time and $g=9.8\ m/s^2$

Furthermore, an object dropped from a certain height h will fall a distance y, given by:

$y=g.t^2/2$

Thus, the height of this object above the ground is:

$H = h-g.t^2/2$

The question describes that ball 1 is dropped from a height of h=22 m. At the same time, ball 2 is thrown straight up with vo=12 m/s.

We want to find at what height both balls coincide. We'll do it by finding the time when it happens. We have written the equations for the height of both balls, we only have to equate them:

$vo.t-g.t^2/2=h-g.t^2/2$

Simplifying:

$vo.t=h$

Solving for t:

$t=h/vo=22/12=1.833\ s$

The height of ball 1 is:

$H = 22-9.8.(1.833)^2/2$

H = 5.53 m

The height of ball 2 is:

$y=12\cdot(1.833)-9.8\cdot(1.833)^2/2$

y=5.53 m

As required, both heights are the same.

The speed of the first ball is:

$vf1=g.t=9.8\cdot 1.833=17.97\ m/s$

vf1=17.97 m/s

The speed of the second ball is:

$vf2=vo-gt=12-9.8\cdot 1.833=-5.96\ m/s$

vf2=-5.96 m/s

This means the second ball is returning to the ground when both balls meet

3 0

3 years ago

A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w

Aloiza [94]

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

4 0

3 years ago

How much heat must be removed from 200 pounds of blanched vegetables at 189 degrees F to freeze them to a temperature of 22 degr

Virty [35]

To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.

By definition the exchange of heat is given by

$Q =mc\Delta T$

where,

m = mass

c = specific heat

$\Delta T$ = Change in temperature

Therefore the total heat exchange is given as

$\Delta Q = Q_w+Q_v$

$\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)$

Our values are given as,

Total mass is $M_T$ = 200lb ,however the mass of solid vegetable and water is given as,

$m_v= 0.4*200lb = 80lb$

$m_w=0.6*200lb=120lb$

$T_i = 183\°F\\T_f = 22\°F\\c_w = 1Btu/lbF\\c_v = 0.24Btu/lbF$

Replacing at our equation we have,

$\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)$

$\Delta Q = (120)(1)(183-22)+(80)(0.24)(183-22)$

$\Delta Q = 22411.2Btu$

Therefore the heat removed is 22411.2 Btu

6 0

3 years ago