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Lubov Fominskaja [6]
3 years ago
5

A tree is fixed relative to Earth a tree is blank relative to the Sun

Physics
1 answer:
RUDIKE [14]3 years ago
8 0

moving,

the sun is a different "observer"

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A swimmer is capable of swimming 0.42 m/s in still water. part a if she aims her body directly across a 66-m-wide river whose cu
satela [25.4K]
<span>If the swimmer is swimming perpendicular to the current, it will take her 66m / 0.42 m/s = 157.14 seconds to cross the river. At the same time, the current will be taking her downstream at a rate of 0.32 m/s. So, when she reaches the opposite bank, her total downstream distance traveled will have been 0.32*157.14 = 50.28 meters.</span>
3 0
3 years ago
You just explored work and heat. Now test your understanding,
Lyrx [107]

Answer:

the hotter it gets, the liquid(mercury), expands more and more, and will rise up the tube to the correct line to read the tempature

5 0
3 years ago
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0
3 years ago
An object with a mass m slides down a rough 370 inclined plane where the coefficient of kinetic friction is 0.20. If the plane i
Svetllana [295]

Answer:

v \approx 9.312\,\frac{m}{s}

Explanation:

The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:

K_{A} + U_{g,A} = K_{B} + U_{g,B} + W_{loss}

K_{B} = K_{A} + U_{g,A}-U_{g,B} - W_{loss}

\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot s\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cdot s \cos \theta

\frac{1}{2}\cdot v^{2} = g\cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)

v = \sqrt{2\cdot g \cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}

v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (10\,m)\cdot (\sin 37^{\textdegree} - 0.2\cdot \cos 37^{\textdegree})}

v \approx 9.312\,\frac{m}{s}

3 0
3 years ago
In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.
Likurg_2 [28]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

Q_H

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

Q_C>0

That is why the answer will be

Q_H ,Q_C>0

8 0
3 years ago
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