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Radda [10]
3 years ago
13

when an object is placed near a concave mirror, at what position does it forms a magnified and erect image.​

Physics
1 answer:
alexira [117]3 years ago
4 0

Answer:

Between the principal focus and the pole of the mirror

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PLEASEEEE HEEEEELP!!!!!
8090 [49]

Answer:

Before:

p_{truck}=16400\ kg.m/s

p_{car}=10000\ kg.m/s

After:

p_{truck}=8000\ kg.m/s

p_{car}=8400\ kg.m/s

v_{fcar}=8.4\ m/s

F=9333.33 \ Nw

Explanation:

<u>Conservation of Momentum</u>

Two objects of masses m1 and m2 moving at speeds v1o and v2o respectively have a total momentum of

p_1=m_1v_{1o}+m_2v_{2o}

After the collision, they have speeds of v1f and v2f and the total momentum is

p_2=m_1v_{1f}+m_2v_{2f}

Impulse J is defined as

J=F.t

Where F is the average impact force and t is the time it lasted

Also, the impulse is equal to the change of momentum

J=\Delta p

As the total momentum is conserved:

p_1=p_2

m_1v_{1o}+m_2v_{2o}=m_1v_{1f}+m_2v_{2f}

We can compute the speed of the second object by solving the above equation for v2f

\displaystyle v_{2f}=\frac{ m_1v_{1o}+m_2v_{2o} -m_1v_{1f} }{  m_2 }

The given data is

m_1=2000\ kg

m_2=1000\ kg\\v_{1o}=8.2\ m/s\\v_{2o}=0\ m/s\\v_{1f}=4\ m/s

a) The impulse will be computed at the very end of the answer

b) Before the collision

p_{truck}=2000\cdot 8.2=16400\ kg.m/s

p_{car}=1000\cdot 0=0\ kg.m/s

c) After collision

p_{truck}=2000\cdot 4=8000\ kg.m/s

Compute the car's speed:

\displaystyle v_{2f}=\frac{ 16400+0 -8000 }{ 1000 }

v_{2f}=8.4\ m/s

And the car's momentum is

p_{car}=1000\cdot 8.4=8400\ kg.m/s

The Impulse J of the system is zero because the total momentum is conserved, i.e. \Delta p=0.

We can compute the impulse for each object

J_1=\Delta p_1=2000(4-8.2)=-8400 \N.s

The force can be computed as

\displaystyle F=\frac{J}{t}=-\frac{8400}{0.9}=-9333.33 \ Nw

The force on the car has the same magnitude and opposite sign

7 0
3 years ago
A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p
anzhelika [568]

Answer:

The final temperature of the air is T_2= 605 K

Explanation:

We can start by doing an energy balance for the closed system

\Delta KE+\Delta PE+ \Delta U = Q - W

where

\Delta KE = the change in kinetic energy.

\Delta PE = the change in potential energy.

\Delta U = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so \Delta KE = 0 and \Delta PE=0

and our energy balance equation is \Delta U = Q - W

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change \Delta U=W+Q using the energy balance equation.

We convert the interval of time to seconds t = 5 \:min = 300\:s

\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t

\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ

We can use the change in specific internal energy \Delta U = m(u_2-u_1) to find the final temperature of the air.

We are given that T_1=300 \:K and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy u_1

u_1=214.07\:\frac{kJ}{kg}

Next, we will calculate the final specific internal energy u_2

\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1

\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}

u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}

With the value u_2=439.07 \:\frac{kJ}{kg} and the ideal gas tables for air we make a regression between the values u = 434.78 \:\frac{kJ}{kg},T=600 \:K and u = 442.42 \:\frac{kJ}{kg}, T=610 \:K and we find that the final temperature T_2 is 605 K.

3 0
3 years ago
Assuming you are in the Northern Hemisphere, how would you expect the location of the Sun in your local sky at noon to differ fr
topjm [15]

Answer:

It should be higher during the summer.

Explanation:

Assuming that youlive in the Northern Hemisphere, the sun's location through the season from winter to summer at noon, should be different. The sun would be observed at a higher position over the horizon in the summer season than it is in the winter season. The reasons are equinoxes, solstices and the ecliptic circle.

I hope this answer helps.

3 0
3 years ago
EMERGENCY! PLS HELP
Karolina [17]

Answer:

0.75 grams

Explanation:

Today we have 6 grams

In 30 years, 3 grams remain

In 60 years 1.5 grams remain

in 90 years 0.75 grams remain

mathematically It would look like this

m = 6(0.5⁽ⁿ/³⁰⁾) where n is number of years

5 0
3 years ago
John's grandfather clock has a pendulum that keeps the seconds. What does this graph say about the observed pattern of motion?
Radda [10]

D. The osculations show a variable rate of motion. Hope this helps:)

7 0
3 years ago
Read 2 more answers
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