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2 years ago

6

What mass of water is required to absorb 4.7 x 10 to the 5th power J of energy from a car engine while the temperature increases

from 298K to 355 K
HELPPPPP

1 answer:

crimeas [40]2 years ago

3 0

I would answer but the other guy got here first and I don’t know the answer sorry

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If you quadruple the temperature of a black body, by what factor will the total energy radiated per second per square meter incr

arlik [135]

<u>In modern physics</u>, as it was called "Stefan-Boltzmann law", the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's temperature T

as:

$P\alpha T^4$

where: P is the power (total energy radiated per second per square meter) and T is the temperature of a black body.

then we can make a ratio between the state of before quadruple (with subscript 1) and after (with subscript 2) as:

$\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{T_{2}^4}$

As

$T_{2}=4T_{1}$

Then

$\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{(4T_{1})^4}$

then

$P_{2}=256*P_{1}$

The factor will the total energy radiated per second per square meter increase = 256

5 0

3 years ago

A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1.5 seconds

horrorfan [7]

Answer:

The force has been reduced by 8018 N

Explanation:

The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:

$F\Delta t = m\Delta v$

where:

F is the force exerted on the car

$\Delta t$ is the duration of the collision

m = 1400 kg is the mass of the car

$\Delta v=-27 m/s$ is the change in velocity of the car

We can re-write the equation as

$F=\frac{m\Delta v}{\Delta t}$

In the 1st collision, the time is 1.5 seconds, so the force is

$F_1=\frac{(1400)(-27)}{1.5}=-25,200 N$

In the 2nd collision, the time is increased to 2.2 seconds, so the force is

$F_2=\frac{(1400)(-27)}{2.2}=-17,182 N$

Therefore, the force has been reduced by:

$F_2-F_1=-17,182-(-25,200)=8018 N$

4 0

3 years ago

Read 2 more answers

In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables

Juli2301 [7.4K]

They made me do it I don’t even know what to say I’m so sorry

7 0

3 years ago

A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The ca

sergeinik [125]

Answer:

t = 8.45 sec

car distance d = 132.09 m

bike distance d = 157.08 m

Explanation:

GIVEN :

motorcycle is 25 m behind the car , therefore distance need to covered by bike to overtake car is 25+ d, when car reache distance d at time t

for car

by equation of motion

$d = ut + \frac{1}{2}at^2$

u = 0 starting from rest

$d = \frac{1}{2}at^2$

$t^2 = \frac{2d}{a}$

for bike

$d+25 = 0 + \frac{1}{2}*4.40t^2$

$t^2= \frac{d+25}{2.20}$

equating time of both

$\frac{2d}{a} = \frac{d+25}{2.20}$

solving for d we get

d = 132 m

therefore t is $= \sqrt{\frac{2d}{a}}$

$t = \sqrt{\frac{2*132}{3.70}}$

t = 8.45 sec

each travelled in time 8.45 sec as

for car

$d = \frac{1}{2}*3.70 *8.45^2$

d = 132.09 m

fro bike

$d = \frac{1}{2}*4.40 *8.45^2$

d = 157.08 m

7 0

3 years ago

I need help with the circled one

Whitepunk [10]

Yes the answer is yes

7 0

3 years ago