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statuscvo [17]
3 years ago
10

According to ohm's law if you don't change the value of the resistor & you double the voltage in a circuit the amount of cur

rent flowing will
A. Be multiplied by the amount of resistance
B. Double
C. Remain the same also
D. Divide it in half
Physics
1 answer:
Nana76 [90]3 years ago
3 0

They double!! Hope im not too late!!

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In an experiment, the 1 kg cart collides with a 3 kg cart but doesn’t stick to it. Instead, the 3 kg cart gets knock forward by
notka56 [123]

Explanation:

Given that,

Mass of the cart, m_1=1\ kg

Mass of the cart 2, m_2=3\ kg            

Final speed of cart 2, v_2=0.3\ m/s

Final speed of cart 1 is 0 as it comes to rest.

Let us assume that the initial velocity of the cart 2 is 0. So using the conservation of linear momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1u_1+0=0+m_2v_2\\\\m_1u_1=m_2v_2\\\\u_1=\dfrac{m_2v_2}{m_1}\\\\u_1=\dfrac{3\times 0.3}{1}\\\\u_1=0.9\ m/s

So, the initial velocity of the 1.0-kg cart is 0.9 m/s.      

8 0
3 years ago
A fuzzy bunny sees a butterfly and chases it right off of a 20 m high cliff at 7m/s
bogdanovich [222]

A) 2.02 s

To find the time it takes for the bunny to reach the rocks below the cliff, we just need to analyze its vertical motion. This motion is a free fall motion, which is a uniformly accelerated motion. So we can use the suvat equation:

s=u_y t+\frac{1}{2}at^2

where

s is the vertical displacement

u is the initial vertical velocity

a is the acceleration

t is the time

For the bunny here, choosing downward as positive direction,

u_y = 0 (initial vertical velocity is zero)

s = 20 m

a=g=9.8 m/s^2 (acceleration of gravity)

And solving for t, we find the time of flight:

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s

B) 14.1 m

For this part, we need to consider the horizontal motion of the bunny.

The horizontal motion of the bunny is a uniform motion with constant velocity, which is the initial velocity of the bunny:

v_x = 7 m/s

Therefore the distance covered after time t is given by

d=v_x t

And substituting the time at which the bunny hits the ground,

t = 2.02 s

We find how far the bunny went from the cliff:

d=(7)(2.02)=14.1 m

C) 21.0 m/s at 70.5^{\circ} below the horizontal

The horizontal component of the velocity is constant during the entire motion, so it will still be the same when the bunny hits the ground:

v_x = 7 m/s

Instead, the vertical velocity is given by

v_y = u_y +at

And substituting t = 2.02 s, we find the vertical velocity at the moment of impact:

v_y = 0+(9.8)(2.02)=19.8 m/s

So, the magnitude of the final velocity is

v=\sqrt{v_x^2+v_y^2}=\sqrt{7^2+(19.8)^2}=21.0 m/s

And the angle is given by

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{19.8}{7})=70.5^{\circ}

below the horizontal

5 0
3 years ago
Describe a situation where you might have high velocity but low acceleration
Dennis_Churaev [7]
In a level cruise on a passenger airliner, flying in a straight line at a steady 550 miles per hour.
Acceleration is zero.
6 0
3 years ago
A 2.0 kg mass attached to an ideal spring oscillates horizontally with an amplitude of 0.15 m. The spring constant is 85N/M
dimulka [17.4K]

Answer:

I think the answer is electric potential.

4 0
3 years ago
A 7.2 m long ramp is set at an angle of 42 degrees with the floor. An 11kg box is pushed up the ramp. What is the change in GPE
Evgesh-ka [11]

Answer:

Answer is 519.3 J

Explanation:

GPE = m x g x h

Object Height is equal to 7.2 x sin42  = 7.2 x 0.669 = 4.817m

GPE = 11 x 9.8 x 4.817 =519.3 J

Most probably m value (11kg) has been used as 7.2 mistakenly.

7 0
4 years ago
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