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3 years ago

1. ____________is defined as the distance an object travels per unit of time.

Physics

1 answer:

charle [14.2K]3 years ago

7 0

Answer:

1) Speed

2) X ( maybe acceleration)

3) Motion

4) Vector

5) Distance

6) Scaler

7) Second

8) Velocity

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The largest tendon in the body, the Achilles tendon, connects the calf muscle to the heel bone of the foot. This tendon is typic

tia_tia [17]

Answer:

Tension on tendon = 1669800N

Explanation:

Detailed explanation and calculation is shown in the image below

7 0

3 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

3 years ago

Read 2 more answers

An electron moves at a speed of 1.0 x 104 m/s in a circular path of radius 2 cm inside a solenoid. The magnetic field of the sol

iogann1982 [59]

Answer:

(a) B = 2.85 × $10^{-6}$ Tesla

(b) I = I = 0.285 A

Explanation:

a. The strength of magnetic field, B, in a solenoid is determined by;

r = $\frac{mv}{qB}$

⇒ B = $\frac{mv}{qr}$

Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field

B = $\frac{9.11*10^{-31*1.0*10^{4} } }{1.6*10^{-19}*0.02 }$

= $\frac{9.11*10^{-27} }{3.2*10^{-21} }$

B = 2.85 × $10^{-6}$ Tesla

b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;

B = μ I N/L

⇒ I = B ÷ μN/L

where B is the magnetic field, μ is the permeability of free space = 4.0 × $10^{-7}$ Tm/A, N/L is the number of turns per length.

I = B ÷ μN/L

= $\frac{2.85*10^{-6} }{4*10^{-7} *25}$

I = 0.285 A

5 0

3 years ago

Please help! Will give thanks and branliest!!!!

slamgirl [31]

<span>The two factors that act on parachutes are gravity and air resistance, which is also called drag. Gravity acts as a force to pull parachutes down to the surface of the Earth, while air resistance generates movement in the opposite direction of the falling parachute, and essentially pushes the parachute upward. hope this helps!:)</span>

6 0

3 years ago

An electric company charges $9.55 per month plus $0.14 per? kilowatt-hour (kwh) for the first 250 kwh used per month and $0.03 p

storchak [24]

$34.75 per month

It is a trick question at the end because it says that anything over 250 kwh is $0.03. Although, you are only calculating for 180 kwh and the monthly charge.

8 0

3 years ago