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2 years ago

3. A cat pushes a 0.25-kg toy with a net force of 8 N. According to Newton's second

Physics

1 answer:

jek_recluse [69]2 years ago

5 0

Mass=0.25kg
Force=8N

$\\ \sf{:}\!\implies F=ma$

$\\ \sf{:}\!\implies Acceleration=\dfrac{F}{m}$

$\\ \sf{:}\!\implies Acceleration=\dfrac{8}{0.25}$

$\\ \sf{:}\!\implies Acceleration=32m/s^2$

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how long would it take for a resultant upward force of 100N to increase the speed of 50kg object from 100m/s to 150m/s

ValentinkaMS [17]

Answer:

25 sec

Explanation:

F = ma

100 = 50 a

a = 2 m/s^2

vf = vo + at

150 = 100 + 2 t

50 = 2 t

t = 25 sec

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1 year ago

A heat engine exhausts 8900 j of heat while performing 2800 j of useful work. part a what is the efficiency of this engine?

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Given: Heat Qout means useful work = 2800 J

Heat Qin = 8900 J

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2 years ago

A force of 6.00 N acts in the positive direction on a 3.00 kg object, originally traveling at +15.0 m/s, for 10.0 s. (a) What is

frozen [14]

Answer:

60 kg m/s

Explanation:

Let $a\;\; m/s^2$ be the acceleration of the object.

As the acceleration of the object is constant, so

$a=\frac {v-u}{t}\cdots(i)$

Given that applied force, F=6.00 N,

From Newton's second law, we have

$F= m\times a$ ,

$\Rightarrow F=\frac {m(v-u)}{t}$ [from equation (i)]

$\Rightarrow Ft=m(v-u)$

$\Rightarrow Ft=mv-mu$

$\Rightarrow mv-mu=6\times 10$ [given that time, t=10 s and F=6 N]

$\Rightarrow mv-mu=60 kg \;m/s$

Here mv is the final momentum of the object and mu is the initial momentum of the object.

So, the change in the momentum of the object is mv-mu.

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2 years ago

A car of 900 kg mass is moving at the velocity of 60 km/hr. It is brought into rest at 50 meter distance by applying a brake. No

kozerog [31]

Answer: $-2502N$

Explanation:

$(V_2)^2=(V_1)^2+2ad$

where;

$V_2$ = final velocity = 0

$V_1$ = initial velocity = 60 km/h = 16.67 m/s

$a$ = acceleration

$d$ = distance

First all of, because acceleration is given in m/s and not km/h, you need to convert 60km/h to m/s. Our conversion factors here are 1km = 1000m and 1h = 3600s

$60km/h(\frac{1000m}{1km} )(\frac{1h}{3600s} )=16.67m/s$

Solve for a;

$(V_2)^2=(V_1)^2+2ad$

Begin by subtracting $(V_1)^2$

$(V_2)^2-(V_1)^2=2ad$

Divide by 2d

$\frac{(V_2)^2-(V_1)^2}{2d} =a$

Now plug in your values:

$a=\frac{(0)^2-(16.67 m/s)^2}{2(50m)}$

$a=\frac{0-277.89m^2/s^2}{100m}$

$a=-2.78m/s$

If you're wondering why I calculated acceleration first is because in order to find force, we need 2 things: mass and acceleration.

$F=ma$

m = mass = 900kg

a = acceleration = -2.78m/s

$F=(900kg)(-2.78m/s)\\F=-2502N$

It's negative because the force has to be applied in the opposite direction that the car is moving.

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2 years ago

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Contact [7]

Answer: Electric field vector is created by an electric force of a charged particle.

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2 years ago