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Studentka2010 [4]

3 years ago

8

A 15.0 kg medicine ball is thrown at a velocity of 10.0 m/s to a 60.0 kg person who is a rest on ice. The person catches the bal

l and immediately slides across the ice. Assume that momentum is conserved and there is no friction. Calculate the velocity of the person AND THE BALL.

2 answers:

shutvik [7]3 years ago

7 0

Answer:

The velocity of ball and man after catching ball is $2m/s$

Explanation:

Since there is no friction and no external force the momentum is conserved .

Here when man catches ball then man and ball move with common velocity .

Let the final common velocity be $v_{f}$ ,

Given that mass of the ball is $m_{b}$ = $15kg$

Given that mass of man is $m=60kg$

Initial velocity of ball = $v_{i} = 10m/s$

Now considering momentum conservation

$p_{i} =p_{f}$

$m_{b}v_{i} = (m_{b} +m)\times v\\v=m_{b}v_{i}/(m_{b} +m) = \frac{150}{75} =2m/s$

Alexxandr [17]3 years ago

7 0

Answer:

Ball = 150kg*m/s

Explanation:

How to find momentum, p:

p = m*v (P=MOMENTUM, M=MASS, V=VELOCITY OF BALL)

P = 15*10

P=150

For the human, momentum is always conserved, so what the momentum is before the ball is thrown must equal after the ball is thrown:

pBEFORE= pAFTER

mv = mv

(15)(10) = (60)(v) SOLVE FOR V

150 = 60v

v = 5/3

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¿A qué velocidad debe circular un auto de carreras para recorrer 87 km en 20 min? (pasar a metros por segundo m/s)

Alexxx [7]

Answer:

Velocidad en m / s = 72,25 m / s

Explanation:

Dado

Distancia a recorrer por el coche de carreras = 87 Km

1 km = 1000 m

Por lo tanto, 87 km = 87000 m

Tiempo necesario para viajar 87 km / 87000 metros = 20 minutos o 20 * 60 = 1200 segundos

Velocidad en m / s = 87000/1200

Velocidad en m / s = 72,25 m / s

7 0

3 years ago

What current is needed to generate the magnetic field strength of 5.0×10−5T at a point 1.5 cm from a long, straight wire? Expres

mixer [17]

Answer:

3.7 A

Explanation:

Parameters given:

Magnetic field strength, B = 5 * 10^(-5) T

Distance of magnetic field from wire, r = 1.5 cm = 0.015 m

The magnetic field, B, due to a current, I, flowing a wire is given as:

B = (μ₀*I) / 2πr

Where μ₀ = permeability of free space

To get the current, I, we make I the subject of the formula:

I = (2πr * B) / μ₀

I = (2 * 3.142 * 5 * 10^(-5)) / (1.25663706 × 10^(-6))

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4 0

3 years ago

Calculate the energy needed to heat 4 kg of water from 25°C to 45°C.

inna [77]

(1 cal/g °C) x (4000 g) x (45 - 25)°C = 80000 cal = 80 kcal. So the answer is 80 kcal .

8 0

3 years ago

Read 2 more answers

When you throw a ball up in the air, it travels up and then stops instantaneously before falling back down. At the point where i

Gnoma [55]

Answer:

The ball stops instantaneously at the topmost point of the motion.

Explanation:

Assume we have thrown a ball up in the air. For that we have given a force on the ball and it acquires an initial velocity in the upward direction.

The forces that resist the motion of the ball in the upward direction are the force of gravity and air resistance. The ball will instantaneously come to rest when the velocity of the ball reduces to zero.

The two forces acting in the downward direction reduces its speed continuously and it becomes zero at the topmost point.

4 0

3 years ago

The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

7 0

3 years ago