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3 years ago

What is the force on a body of mass 20 kg moving with an acceleration of 3 m/s2 ?

Physics

2 answers:

Ahat [919]3 years ago

7 0

Answer:

$\boxed{Force \: on \: body = 60 \: N}$

Given:

Mass of body = 20 kg

Acceleration = 3 m/s²

Explanation:

Force = mass × acceleration

= 20 × 3

= 60 kg m/s² or 60 N

zubka84 [21]3 years ago

5 0

Hey!!

Solution,

Mass(m)=20kg

Acceleration due to gravity(g)=3 m/s^2

Force=?

Now,

Force=m*g

 = 20*3

 = 60 Newton

The force is 60 Newton 

Hope it helps

Good luck on your assignment..

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You throw a basketball and a tennis ball across the classroom so that each ball has the exact same momentum. Explain how this is

Helga [31]

Answer:

It is possible by increasing the speed of the tennis ball by a factor of (Mass of the tennis ball)/(Mass of the basketball)

Explanation:

The momentum of a body = The bod's mass × The body's velocity

Therefore, the momentum of a given mass of an object, such as a tennis ball can be increased by increasing the velocity or speed of the object. Whereby the speed of the ball, v₁, is increased such that the momentum of the basketball and the tennis ball will be the same, is given by the following equation

Mass of the basketball × v₂ = Mass of the tennis ball × v₁

Therefore, v₁/v₂ = (Mass of the tennis ball)/(Mass of the basketball)

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2 years ago

A ball on a string makes 25.0 revolutions in 9.37 s, in a circle radius 0.450 m. What is it’s velocity?

Sauron [17]

The velocity of the ball is 12.5 m/s

Explanation:

The velocity of the ball is given by the ratio between the distance covered by the ball and the time taken:

$v=\frac{d}{t}$

First, we calculate the distance covered. We know that the radius of the circle is

r = 0.450 m

And the length of the circumference is

$L=2\pi r = 2\pi(0.750)=4.7 m$

The ball makes 25.0 revolutions, so a total distance of

$d=(25.0)L=(25.0)(4.7)=117.5 m$

In a time of

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Learn more about velocity here:

brainly.com/question/5248528

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2 years ago

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Air, considered an ideal gas, is contained in an insulated piston-cylinder assembly outfitted with a paddle wheel. It is initial

Maru [420]

Our data are,

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$P_1= 10psi=68.95kPa\\V_1 = 1ft^3=0.02831m^3\\T_1 = 100\°F = 310.93K$

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$P_2 =5psi=34.474kPa\\V_2 = 3ft^3=0.0899m^3$

We know as well that $3BTU=3.16kJ/K$

To find the mass we apply the ideal gas formula, which is given by

$P_1V_1=mRT_1$

Re-arrange for m,

$m= \frac{P_1V_1}{RT_1}\\m= \frac{68.95*0.02831}{(0.287)310.9}\\m=0.021893kg=0.04806lbm\\$

Because of the pressure, temperature and volume ratio of state 1 and 2, we have to

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Replacing,

$T_2 = \frac{P_2V_2}{P_1V_1}T_1\\T_2 =\frac{34.474*0.0844}{68.95*0.02831}*310.93\\T_2 = 464.217K=375.5\°F$

For conservative energy we have, (Cv = 0.718)

$W = m C_v = 0.718 \Delta T +dw\\dw = W - mv\Delta T\\dw = 3.16-(0.0218*0.718)(454.127-310.93)\\dw = 0.765kJ=0.72BTU$

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2 years ago

Can anyone solve these for my by using unit vectors? Can you also please show your work

Oxana [17]

4. The Coyote has an initial position vector of $\vec r_0=(15.5\,\mathrm m)\,\vec\jmath$ .

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$\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2$

where $\vec a$ is the Coyote's acceleration vector at time $t$ . He experiences acceleration only in the downward direction because of gravity, and in particular $\vec a=-g\,\vec\jmath$ where $g=9.80\,\frac{\mathrm m}{\mathrm s^2}$ . Splitting up the position vector into components, we have $\vec r=r_x\,\vec\imath+r_y\,\vec\jmath$ with

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t$

$r_y=15.5\,\mathrm m-\dfrac g2t^2$

The Coyote hits the ground when $r_y=0$ :

$15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s$

4b. Here we evaluate $r_x$ at the time found in (4a).

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m$

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$r_y=1.52\,\mathrm m-\dfrac g2t^2$

We find the shell hits the ground at

$1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s$

5b. The horizontal component of the bullet's position vector is

$r_x=v_0t$

where $v_0$ is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for $v_0$ :

$3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}$

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3 years ago

an object weighs 98 n on earth. How much does it weigh on planet x where the acceleration due to gravity in 6 m/s^2

Degger [83]

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3 years ago

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