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rodikova [14]
3 years ago
7

A high-jumper clears the bar and has a downward velocity of - 5.00 m/s just before landing on an air mattress and bouncing up at

1.0 m/s. The mass of the high-jumper is 60.0 kg. What is the magnitude and direction of the impulse that the air mattress exerts on her
Physics
1 answer:
Jobisdone [24]3 years ago
8 0

-- As she lands on the air mattress, her momentum is (m v)

Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down

-- As she leaves it after the bounce,

Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up

-- The impulse (change in momentum) is

Change = (60 kg-m/s up) - (300 kg-m/s down)

Magnitude of the change = <em>360 km-m/s </em>

The direction of the change is <em>up /\ </em>.

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A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotat
AleksandrR [38]

Answer:

\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}

Explanation:

As we know that there is no external torque on the system of two disc

then the angular momentum of the system will remains conserved

So we will have

L_i = L_f

now we have

L_i = (\frac{1}{2}MR^2)\omega_o

also we have

L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega

now from above equation we have

(\frac{1}{2}MR^2)\omega_o  = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega

now we have

\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}

\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}

6 0
3 years ago
A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00
gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as:W=Q=Q_{in}-Q_{out} solving to Q_out we get:Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ) this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: n=1-\frac{T_{Low} }{T_{High}} where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:n=1-\frac{423.15}{723.15} =0.415=41.5%

5 0
2 years ago
You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar
LuckyWell [14K]

Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

Thus, the expression for final pressure in the two containers will be:

PV=P_1V_1+P_2V_2

P=\frac{P_1V_1+P_2V_2}{V}

where,

P_1 = pressure of N₂ gas = 4.45 atm

P_2 = pressure of Ar gas = 2.75 atm

V_1 = volume of N₂ gas = 3.00 L

V_2 = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}

P=2.62atm

Thus, the final pressure in the two containers is, 2.62 atm

8 0
2 years ago
I know how the voltage increasing/not enough/stays the same and how long the battery lasts with series and parallel connections.
kirill [66]

Answer:

It comes out the positive side of the battery and goes in to the negative side of the battery

Explanation:

There are already electrons in wires in a circuit before you add the battery. By adding the battery, you're giving the electrons the energy it needs to move along the circuit.

In a series circuit, the circuit is one continuous loop so there is only one path for the electrons to go - out of the positive side of the battery and around the circuit then goes back into the negative side of the battery.

However, with a parallel circuit, there are two or more ways the electrons can go so they take the path of least resistance. The electrons still go out the positive side of a battery but along the circuit, the electrons will go through the path of least resistance ( I tend to think of it like a net with holes in it - the lower the resistance the bigger the holes for the electrons to go through so more can fit in a set amount of time ) but the electrons still go out of the positive side and in through the negative

3 0
2 years ago
A wave has a wavelength of 0.98 m and travels 2.0x10^2 m in 0.78 s. What is the frequency of the wave?
asambeis [7]

Answer:

261.64 hz

Explanation:

speed = distance/ time

s= 200/0.78= 256.41 ms^-1

by using v= (f) x ( lambda)

256.41/ 0.98 = f = 261.64 hz

4 0
2 years ago
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