What is the magnification of an object that is 4.15 m in front of a camera that has an image position of 5.0 cm?

A 0.80-kg soccer ball experinces an impulse of 25 N x s . Determine the momentum change of the soccer ball.

borishaifa [10]

If the impulse is 25 N-s, then so is the change in momentum.
The mass of the ball is extra, unneeded information.

Just to make sure, we can check out the units:

Momentum = (mass) x (speed) = kg-meter / sec

Impulse = (force) x (time) = (kg-meter / sec²) x (sec) = kg-meter / sec

3 0

3 years ago

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A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder

Stella [2.4K]

Answer:

The temperature change per compression stroke is 32.48°.

Explanation:

Given that,

Angular frequency = 150 rpm

Stroke = 2.00 mol

Initial temperature = 390 K

Supplied power = -7.9 kW

Rate of heat = -1.1 kW

We need to calculate the time for compressor

Using formula of compression

$\terxt{time for compression}=\text{time for half revolution}$

$\terxt{time for compression}=\dfrac{1}{2}\times T$

$\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{f}$

Put the value into the formula

$\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{150}\times60$

$\terxt{time for compression}=0.2\ sec$

We need to calculate the rate of internal energy

Using first law of thermodynamics

$U=Q-W$

$\dfrac{\Delta U}{\Delta t}=\dfrac{\Delta Q}{\Delta t}-\dfrac{\Delta W}{\Delta t}$

Put the value into the formula

$\dfrac{\Delta U}{\Delta t}=(-1.1)-(7.9)$

$\dfrac{\Delta U}{\Delta t}=6.8\ kW$

We need to calculate the temperature change per compression stroke

Using formula of rate of internal energy

$\dfrac{\Delta U}{\Delta t}=\dfrac{nc_{v}\Delta \theta}{\Delta t}$

$\Delta\theta=\dfrac{\Delta U}{\Delta t}\times\dfrac{\Delta t}{n\times c_{c}}$

Put the value into the formula

$\Delta \theta=6.8\times10^{3}\dfrac{0.2}{2.0\times20.93}$

$\Delta\theta=32.48^{\circ}$

Hence, The temperature change per compression stroke is 32.48°.

6 0

3 years ago

A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby has a mass of 4kg. The carriage ha

Ira Lisetskai [31]

Answer:

E=252J

Explanation:

The total mechanical energy of an object or system is given by:

E mech=K+U

Where K is the kinetic energy of the object and U is the potential energy of the object. The carriage, sitting motionless at the top of the hill, has only potential energy in the form of gravitational potential energy.

Gravitational potential energy is given by:

Ug=mgh

Where m is the mass of the object, g is the gravitational acceleration constant, and h is the height of the object above some specific reference point, in this case the ground 21 m below.

The weight of a stationary object at the surface of the earth is equal to the force of gravity acting on the object.

W=→Fg=mg

We are given that the carriage weighs 12 N, therefore mg=12N.

Ug=12N⋅21m

⇒Ug=252Nm=252J

Hope it helped, God bless you!

5 0

3 years ago

A bullet with mass 1.0kg and velocity 180 m/s is brought to rest in 0.02 s by a sandbag.assuming constant acceleration in the sa

kotykmax [81]

Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is $v_i=180~m/s$ , the final velocity is $v_i=0~m/s$ , and the total time of the motion is $\Delta t=0.02~s$ , so the acceleration is given by
$a= \frac{v_f-v_i}{\Delta t} = -9000~m/s^2$
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
$S=v_i t + \frac{1}{2}at^2 = 180~m/s \cdot 0.02~s + \frac{1}{2}(-9000~m/s^2)(0.02~s)^2=1.8~m$
So, the bullet penetrates the sandbag 1.8 meters.

5 0

3 years ago

Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up

Mademuasel [1]

Answer:

V = 90.51 m/s

Explanation:

From the given information:

Initial speed (u) = 0

Distance (S) = 391 m

Acceleration (a) = 18.9 m/s²

Using the relation for the equation of motion:

v² - u² = 2as

v² - 0² = 2as

v² = 2as

$v = \sqrt{2as}$

$v = \sqrt{2*18.9*391}$

v = 121.57 m/s

After the parachute opens:

The initial velocity = 121.57 m/ss

Distance S' = 332 m

Acceleration = -9.92 m/s²

How fast is the racer can be determined by using the relation:

$V= \sqrt{v^2 + 2aS'}$

$V = \sqrt{121.57^2+ 2 (-9.92)(332)}$

V = 90.51 m/s

6 0

3 years ago