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3 years ago

Tim observes that a force pulls him down when he jumps up in the air. Based on his observation, which conclusion is correct?

Physics

1 answer:

enyata [817]3 years ago

8 0

Gravitational force is a non-contact force.

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How does the structure of mitochondrial on affect its structure

anyanavicka [17]

It is itself. This question does not make sense.

7 0

3 years ago

A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th

Evgen [1.6K]

Answer:

$\dfrac{d\theta}{dt}=0.038\ rad/s$

Explanation:

Given that

$\dfrac{dx}{dt}= -1\ m/s$

From the diagram

$tan\theta=\dfrac{21}{x}$

By differentiating with time t

$sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}$

When x= 10 m

$tan\theta=\dfrac{21}{10}$

θ = 64.53°

Now by putting the value in equation

$sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}$

$sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)$

$\dfrac{d\theta}{dt}=0.038\ rad/s$

Therefore rate of change in the angle is 0.038\ rad/s

8 0

3 years ago

A comparison of a machine’s work output with its work input is called __??__.

Aloiza [94]

Ohhhhh its called a input machine

7 0

3 years ago

A projectile is shot from the edge of a cliff 80 m above ground level with an initial speed of 60 m/sec at an angle of 30° with

Dvinal [7]

Answer:

8 seconds

Explanation:

Answer:

Explanation:

Going up

Time taken to reach maximum height= usin∅/g

=3 secs

Maximum height= H+[(usin∅)²/2g]

=80+[(60sin30)²/20]

=125 meters

Coming Down

Maximum height= ½gt²

125= ½(10)(t²)

t=5 secs

6 0

3 years ago

An inventor claims to have developed a power cycle having a thermal efficiency of 40%, while operating between hot and cold rese

lesya [120]

From Carnot's theorem, for any engine working between these two temperatures:

efficiency <= (1-tc/th) * 100

Given: tc = 300k (from question assuming it is not 5300 as it seems)

For a, th = 900k, efficiency = (1-300/900) = 70%

For b, th = 500k, efficiency = (1-300/500) = 40%

For c, th = 375k, efficiency = (1-300/375) = 20%

Hence in case of a and b, efficiency claimed is lesser than efficiency calculated, which is valid case and in case of c, however efficiency claimed is greater which is invalid.

7 0

3 years ago