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2 years ago

A person throws a ball with a force of 140 N that accelerates at 15 m/s? What is the mass of the

Physics

1 answer:

mr_godi [17]2 years ago

6 0

Newton's 2nd law

$\tt \sum F=m.a$

input the value:

$\tt m=\dfrac{\Sum F}{a}=\dfrac{140~N}{15~m/s^2}=9.3~kg$

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A catcher in a baseball game stops a pitched ball that was originally moving at 44 m/s over a distance of 12.5 cm. the mass of t

eduard

Answer:

Force = -1161.6 Newton

Explanation:

Given the following data;

Initial velocity, u = 44m/s

Distance ,s = 12.5cm to m = 12.5/100 = 0.125m

Mass = 0.15kg

To find the acceleration;

We would use the third equation of motion;

V ² = U² + 2as

0² = 44² + 2*a*0.125

0 = 1936 + 0.25a

0.25a = -1936

a = -1936/0.25

Acceleration, a = -7744m/s2

Force = mass * acceleration

Substituting into the equation, we have;

Force = 0.15 * (-7744)

Force = -1161.6 Newton

The value of its force is negative because the glove decreases the velocity of the ball.

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2 years ago

Write the following as powers of ten with one figure before the decimal point:

vodka [1.7K]

........................z......

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2 years ago

An uncrewed mission to the nearest star, Proxima Centauri, is launched from the Earth's surface as a projectile with an initial

Anna [14]

Answer:

42.96 km/s

Explanation:

From the conservation of Energy

$(PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2$

Mass gets cancelled

$-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}$

$v_e=\sqrt{\frac{2Gm}{R}}$ = Escape velocity of Earth = 11.2 km/s

$v_i$ = Velocity of projectile = 44.4 km/s

$v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s$

The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s

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2 years ago

If the car’s speed decreases at a constant rate from 64 mi/h to 30 mi/h in 3.0 s, what is the magnitude of its acceleration, ass

mixas84 [53]

Answer: $3.874 m/s^2$

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

$60mi/h \approx 26.8224m/s$

$34mi/h \approx 15.1994 m/s$

we know acceleration is given by $=\frac{velocity}{Time}$

$a=\frac{15.1994-26.8224}{3}$

$a=-3.874 m/s^2$

negative indicates that it is stopping the car

Distance traveled

$v^2-u^2=2as$

$\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s$

$s=\frac{488.419}{2\times 3.874}$

s=63.038 m

7 0

2 years ago

A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exe

Karolina [17]

Answer:

The velocity is 40 ft/sec.

Explanation:

Given that,

Force = 3200 lb

Angle = 30°

Speed = 64 ft/s

The resistive force with magnitude proportional to the square of the speed,

$F_{r}=kv^2$

Where, k = 1 lb s²/ft²

We need to calculate the velocity

Using balance equation

$F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}$

Put the value into the formula

$3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}$

Put the value of k

$3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}$

$1600-v^2=m\dfrac{d^2v}{dt^2}$

At terminal velocity $\dfrac{d^2v}{dt^2}=0$

So, $1600-v^2=0$

$v=\sqrt{1600}$

$v=40\ ft/sec$

Hence, The velocity is 40 ft/sec.

4 0

2 years ago