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2 years ago

Why is figure 5 an unhelpful visualization tool for this data set? Please help!

Physics

1 answer:

Paraphin [41]2 years ago

8 0

Explanation:

Because the temperature and the radiation are not correlated, they're not represented as functions of each other, they're represented as independent variables thus using graph 5 you cannot figure out how one affect another

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A dart with mass md is launched toward a block of mass mb that is suspended from a string of length L. The dart is moving horizo

Yuki888 [10]

Answer:

A) Impulse is the same for both the objects

B) The higher is the speed, the greater will be the height.

Explanation:

Part a)

The time of interaction of the two bodies i.e the hanging mass and the stick is same. Thus, force caused by dart on the block = force caused by block on the dart. Hence, impulse is the same for both the objects.

Part B

The energy will be conserved in the entire reaction process

Hence, Kinetic energy = potential energy

0.5Mv^2 = gh(md+mb)

H is directly proportional to the square of speed.

Hence, the higher is the speed, the greater will be the height.

5 0

2 years ago

I). Mechanical energy is the sum of potential energy and kinetic energy in an object that is used to do work.

Dmitry_Shevchenko [17]

Answer:

false statement : b ) For the motion of a cart on an incline plane having a coefficient of kinetic friction of 0.5, the magnitude of the change in kinetic energy equals the magnitude of the change in gravitational potential energy

Explanation:

mechanical energy = potential energy + kinetic energy = constant

differentiating both side

Δ potential energy + Δ kinetic energy = 0

Δ potential energy = - Δ kinetic energy

first statement is true.

Friction is a non conservative force so inter-conversion of potential and kinetic energy is not possible in that case. In case of second option, the correct relation is as follows

change in gravitational potential energy = change in kinetic energy + work done against friction .

So given 2 nd option is incorrect.

In case of no change in gravitational energy , work done is equal to

change in kinetic energy.

4 0

3 years ago

A student at a window on the second floor of a dorm sees her physics professor walking on the sidewalk beside the building. she

klasskru [66]

Refer to the diagram shown below.

In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s

The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m

Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.

The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s

The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.

Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.

8 0

3 years ago

What would you predict for this elephant's stride frequency? That is, how many steps per minute will the elephant take?

Sonja [21]

<span>` You can consider T to be in units of seconds/step. Frequency is the inverse of period, so
1/T = frequency and has units of steps per second. There will be 60 times as many steps in a minute.</span>

8 0

3 years ago

Read 2 more answers

After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg

GuDViN [60]

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses $m_1$ and $m_2$ whose respective velocities before collision are $u_1$ and $u_2$ ;