1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Anna71 [15]
4 years ago
7

An____ collision occurs when an arrow hits the target.

Physics
1 answer:
MrMuchimi4 years ago
8 0

Answer:

I think it's An <u>inelastic</u> collision occurs when an arrow hits the target because when the arrow hits it, the target and the arrow do not bounce away from each other.

Explanation:

You might be interested in
The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr
Amanda [17]

Answer:

The tube should be held vertically and perpendicular to the ground.

Explanation:

Answer: The tube should be held vertically and perpendicular to the ground. The reason is as follows:

Reasoning:

The power lines are parallel to the ground hence, their electric field will be perpendicular to the ground and equipotential surface will be cylindrical.

Hence, if you will put fluorescent tube parallel to the ground then both the ends of the tube will lie on the same equipotential surface and the potential difference will be zero.

So, to maximize the potential the ends of the tube must be on different equipotential surfaces. The surface which is near to the power line has high potential value and the surface which is farther from the line has lower potential value.

hence, to maximize the potential difference, the tube must be placed perpendicular to the ground.

6 0
3 years ago
A 6- F capacitor is charged to 90 V and is then connected across a 700- resistor. What is the initial charge on the capacitor
Lorico [155]

Answer:

540C.

Explanation:

A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;

Q = CV          ----------(i)

From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.

Where;

C = 6F

V = 90V

Substitute these values into equation (i) as follows;

Q = 6 x 90

Q = 540 C

Therefore, the initial charge on the capacitor is 540C.

7 0
3 years ago
Suppose Group A runs their experiment 3 times and calculates that the best estimate of the distance slid by red blocks is 15 + 3
Alex73 [517]
Sorry I just need points
7 0
3 years ago
A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball
adell [148]

Answer:

Part a)

T = 2\sqrt{\frac{R}{3g}}

Part b)

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

v_y = \sqrt{Rg/3}

Part d)

v = \frac{1}{2}\sqrt{13Rg}

Part e)

\theta_i = 33.7 degree

Part f)

H = \frac{13R}{8}

Part g)

X = \frac{13R}{4}

Explanation:

Initial speed of the launch is given as

initial speed = v_i

angle = \theta_i degree

Now the two components of the velocity

v_x = v_i cos\theta_i

similarly we have

v_y = v_i sin\theta_i

Part a)

Now we know that horizontal range is given as

R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}

maximum height is given as

H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}

so we have

v_i sin\theta = \sqrt{Rg/3}

time of flight is given as

T = \frac{2v_isin\theta_i}{g}

T = \frac{2\sqrt{Rg/3}}{g}

T = 2\sqrt{\frac{R}{3g}}

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

R = v_x T

R = v_x 2\sqrt{\frac{R}{3g}}

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

Initial vertical velocity is given as

v_y = v_i sin\theta_i

v_i sin\theta = \sqrt{Rg/3}

Part d)

Initial speed is given as

v = \sqrt{v_x^2 + v_y^2}

so we will have

v = \sqrt{Rg/3 + 3Rg/4}

v = \frac{1}{2}\sqrt{13Rg}

Part e)

Angle of projection is given as

tan\theta_i = \frac{v_y}{v_x}

tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}

\theta_i = 33.7 degree

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

H = \frac{v^2}{2g}

H = \frac{13R}{8}

Part g)

For maximum range the angle should be 45 degree

so maximum range is

X = \frac{v^2}{g}

X = \frac{13R}{4}

3 0
4 years ago
Lab report 1.04 hypothesis
Zanzabum

Answer:a supposition or proposed explanation made on the basis of limited evidence as a starting point for further investigation.

Hope This helps!!

5 0
3 years ago
Other questions:
  • Describe each class of lever and explain to characteristics of each
    5·2 answers
  • How can tectonic activity change the climate of the earth?
    12·1 answer
  • The blackbody curve of a star moving toward Earth would have its peak shifted (a) to lower intensity; (b) toward higher energies
    7·1 answer
  • How can a bicycles kinetic energy change when it slows down?
    11·2 answers
  • According to Bode’s Law a planet is missing between Jupiter and Saturn. True False
    5·1 answer
  • Give a strong idea to protect , purify and the beautify of the Dal lake​
    6·1 answer
  • Please helppppppppppp​
    5·1 answer
  • 17
    13·1 answer
  • A phonograph turntable rotating at 33 1 3 rev/min slows down and stops in 1.98 min. (Assume the turntable initially rotates in t
    8·1 answer
  • If a Ferrari, with an initial velocity of 14m/s, accelerates at a rate of 60
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!