The picture shows an object resting on a balance.

A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra

lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

$m_{1}=5.8 kg$ is the mass of the bowling ball

$V_{1}=4.34 m/s$ is the velocity of the bowling ball

$m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg$ is the mass of the ping-pong ball

$V_{2}$ is the velocity of the ping-pong ball

Now, the momentum $p_{1}$ of the bowling ball is:

$p_{1}=m_{1}V_{1}$ (1)

$p_{1}=(5.8 kg)(4.34 m/s)$

$p_{1}=25.172 kg m/s$ (2)

And the momentum $p_{2}$ of the ping-pong ball is:

$p_{2}=m_{2}V_{2}$ (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

$p_{1}=p_{2}$ (4)

$m_{1}V_{1}=m_{2}V_{2}$ (5)

Isolating $V_{2}$ :

$V_{2}=\frac{m_{1}V_{1}}{m_{2}}$ (6)

$V_{2}=\frac{25.172 kg m/s}{0.002214 kg}$ (7)

Finally:

$V_{2}=11369.46 m/s$

6 0

3 years ago

One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what

Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

$F_y=ma_y$ (2)

$F_x=ma_x$ (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

$F_y=W+n=0$

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

$F_y=(755)(-9.81)+n=0$

$n=7406.55 N$ (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

$F_x=F+f=ma_x$

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

$1988+f=m(1.36)$

$f=(755)(1.36)-1988=-961.2 N$

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

$|f|=\mu_k n$

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

$\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13$

4 0

3 years ago

6.All wheelchairs may be secured so that the user is facing the curb side of the vehicle.

Nataly_w [17]

All wheelchairs may be secured so that the user is facing the curb side of the vehicle is true. The answer is letter A. It provides a unique 180 degree powered rotation which makes it possible to raise, lower and rotate fully.

6 0

3 years ago

Approximate the Sun as a uniform sphere of radius 6.96 X 108 m, rotating about its central axis with a period of 25.4 days. Supp

In-s [12.5K]

Answer:

T = 184 seconds

Explanation:

First in order to solve this, we need to know which is the expression to calculate the period. This is an exercise of angular velocity, so:

T = 2π/w

Where w: angular speed (in rad/s)

So, let's calculate first the innitial angular speed:

w = 2π/T

Converting days to seconds:

25.4 days * 24 h/day * 3600 s/h = 2,194,560 s

Then the angular speed:

w = 2π / 2,194,560 = 2.863x10^-6 rad/s

Now, the innitial angular momentum is:

I = (2/5)Mr² replacing data:

I = 2/5* (6.96x10^8)² * M = 1.94x10^17m² * M

so the initial angular momentum would be:

L = Iω = 2.863x10^-6 * 1.94x10^17 M

L = 5.55x10^11 m²/s * M = final angular momentum

Now the final I = 2/5Mr²

Final I = 2/5 * (6.37x10^6)² * M = 1.62x10^13m² * M

Then 5.55x10^11m²/s * M = 1.62x10^13m² * M * ω → M cancels

ω = 3.42x10^-2 rad/s

Then the new period

T = 2π/ω = 2*3.14 / 3.42x10^-2

T = 184 seconds

8 0

3 years ago

. If measurements of a gas are 75L and 300 kilopascals and then the gas is measured a second time and found to be 50L, describe

julia-pushkina [17]

By Boyle's law:

P₁V₁ = P₂V₂

300*75 = P₂*50

P₂*50= 300*75

P₂ = 300*75/50 = 450

P₂ = 450 kiloPascals.

The pressure has increased as a result of compression of gas.

Boyle's Law supports this observation.

3 0

3 years ago

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