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timurjin [86]
3 years ago
12

If you push on an object that has a mass of 400 kg and it accelerates at 4 m/s2, what would the acceleration (a) be if the mass

was reduced to 1/4 as much?
(Hint: Instead of plugging 400 kg into the equation, plug 100 kg into the equation as the mass but keep the force at 1,600 N.)

a
1/4 m/s2

b
16 m/s2

c
8 m/s2

d
4 m/s2
Physics
1 answer:
Temka [501]3 years ago
6 0

Answer:

b  16 m/s².

Explanation:

Hello.

In this case, the force exerted to the mass of 400 kg at an acceleration of 4 m/s² is:

F=400kg*4m/s^2=1600N

Now, since the mass is decreased to 100 kg (1/4 of 400 kg), and the force remain the same, the new acceleration should be:

a=\frac{F}{m}=\frac{1600kg*m/s^2}{100kg}\\  \\a=16m/s^2

Thus, the answer is b  16 m/s².

Best regards.

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Solve the science problem
scZoUnD [109]

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nolan ryan has the record for having the speediest fastball in baseball. he could pitch one at 148 i/sec. what is that speed in
Klio2033 [76]

The speed of the ball is 101miles/hr.

A mile is a unit of length that is exactly 1,609.344 metres long. Similarly, 5,280 feet or 1,760 yards make up one mile. The mile is an imperial and common US measurement of distance.

We just have to deal with unit conversions.

One mile is 5280 feet, or  1 ft = 0.000189

The speed of the ball in miles per hour is

\frac{148ft}{1sec} . \frac{1mile}{5280ft} .\frac{60s}{1min} .\frac{60min}{1hr}

So, the speed of the ball in miles per hour is 101miles/hr.

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brainly.com/question/23245414

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Two long, straight wires, one above the other, are seperated by a distance 2a2a and are parallel to the x−axisx−axis. Let the +y
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Answer:

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3 years ago
Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem
statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

x=x_o+v_1t    (1)

xo = 10 km

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car 2:

x'=v_2t    (2)

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For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

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The position in which both cars coincides is:

x=(55km/h)(0.33h)=18.33km

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3 years ago
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