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2 years ago

If you run at 1.7 m/s FORWARD ,how does this affect the speed of a ball that you throw?

1 answer:

8 0

We have a problem about conservation and velocity, we will find that it does affect the speed of the ball, increasing it by 1.7m/s.

There is something called momentum, which we can define as the "quantity of movement", and we can simply write as the product between velocity and mass.

The momentum is conservative, then we have conservation of momentum.

This means that when you run whit the ball in your hands, the momentum of the ball will be equal to your velocity times the mass of the ball, and this must conserve after you throw the ball.

Now with this idea in mind, this means that if you run with a velocity V, and you throw the ball with a velocity V', the velocity of the ball when it leaves your hand will be:

V + V'.

So, if you run with a velocity of 1.7m/s forward and you throw the ball (assuming in the same direction) the speed of the ball will be 1.7m/s larger than if you were to throw it standing still.

If you want to learn more, you can read:

brainly.com/question/13639113

You might be interested in

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

3 years ago

When the force acting on the body equal to acceleration?

topjm [15]

Answer:

Acceleration and velocity Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate. In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force.

5 0

2 years ago

What is the gravitational potential energy of a 0.550-kg projectile flying with 335 m/s, 72 meters above the ground?

Fofino [41]

Answer:

GPE = 388.08 Joules.

Explanation:

Given the following data;

Mass = 0.550kg

Speed = 335 m/s

Height = 72 meters

We know that acceleration due to gravity, g is equal to 9.8 m/s²

To find the gravitational potential energy;

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

$G.P.E = mgh$

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the formula, we have;

$G.P.E = 0.550 * 9.8 * 72$

GPE = 388.08 Joules.

4 0

3 years ago

A 400g sample of water absorbs 500j of energy. how did the water temperature change if the specific heat of water is 4.18j/g©. S

Mashutka [201]

In general, the quantity of heat energy, Q, required to raise a mass m kg of a substance with a specific heat capacity of c J/(kg °C), from temperature t1 °C to t2 °C is given by:

Q = mc(t2 – t1) joules

So:

(t2-t1) =Q / mc

As we know:
Q = 500 J

m = 0.4 kg

c = 4180 J/Kg °c

We can take t1 to be 0°c

t2 - 0 = 500 / ( 0.4 * 4180 )

t2 - 0 = 0.30°c

6 0

3 years ago

The volume flow rate in an artery supplying the brain is 4.00 10-6 m3/s. (a) If the radius of the artery is 4.50 10-3 m, determi

mr_godi [17]

Answer:

(a) 0.063 m/s

(b) 1.01 m/s

Explanation:

rate of volume flow, V = 4 x 10^-6 m^3/s

(a) radius, r = 4.5 x 10^-3 m

Let the speed of blood is v.

So, V = A x v

where A be the area of crossection of artery

4 x 10^-6 = 3.14 x 4.5 x 10^-3 x 4.5 x 10^-3 x v

v = 0.063 m/s

Thus, the speed of flow of blood is 0.063 m/s .

(b) Now r' = r / 4 = 4.5 /4 x 10^-3 m = 1.125 x 10^-3 m

Let the speed is v'.

So, V = A' x v'

4 x 10^-6 = 3.14 x 1.125 x 10^-3 x 1.125 x 10^-3 x v'

v' = 1.01 m/s

Thus, the speed of flow of blood is 1.01 m/s .

8 0

3 years ago