Use the equation V=IR to find resistance, potential difference or current given the other two quantities.

Ad libitum [116K]

Hey!!

here is your answer >>>

Specific heat is the amount of heat required to raise the temperature of a unit substance!. The most common example!. The specific heat of water is 1 !. And, lot of plays and animals live under the water and if the specific heat of the water would be low, then the aquatic organisms would die, but the specific heat is high and it is hard for the sun to increase the temperature!.

Hope my answer helps!

3 0

3 years ago

A uniform bridge, 20 m long and weighing 40000 N is supported by two pillars located 3.0 m from each end. If a 19600 N car is pa

vodomira [7]

Let the forces on two pillars are N1 and N2

so we will have force balance equation as

$N_1 + N_2 = 40000 + 19600$

$N_1 + N_2 = 59600 N$

also by torque balance we will have

$N_2\times L = 40000 ]times \frac{L}{2} + 19600 \times 8$

here we know that

L = 20 m

$N_2 (20 m) = 40000 (10 m) + 19600 (8 m)$

$N_2 = 27840 N$

$N_1 = 59600 - 27840 = 31760 N$

8 0

3 years ago

Two identical balls collide one was initially at rest. after the collision one of the balls has a velocity of 3 m/s south. the o

Butoxors [25]

Two identical balls collide head on. The initial velocity of one is 0.75 m/s east, while that of the other one is 0.43 m/s west.

4 0

3 years ago

A boy throws a ball straight up with a speed of 21.5 m/s. The ball has a mass of 0.19 kg. How much gravitational potential energ

astra-53 [7]

Answer:

Explanation:

The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation

$v_f=v_0+at$ where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...

0 = 21.5 + (-9.8)t and

-21.5 = -9.8t so

t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)

Now we will use that time to find out the max height of the object in the equation

Δx = $v_0t+\frac{1}{2}at^2$ and filling in:

Δx = $21.5(2.19)+\frac{1}{2}(-9.8)(2.19)^2$ which simplifies down a bit to

Δx = 47.1 - 23.5 so

Δx = 23.6 meters.

Now we can plug that in to the PE equation to find the PE of the object:

PE = (.19)(9.8)(23.6) so

PE = 43.9 J

5 0

3 years ago

outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and comp

sweet [91]

Answer:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

$v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{85\times (0.065)^2}{1.5}} \\\\v=0.489\ m/s$

So, the speed of the box is 0.489 m/s.

3 0

4 years ago