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Alisiya [41]
3 years ago
15

A roofer drops a nail that hits the ground traveling at 26 m/s. How fast was the nail traveling 1 second before it hits the grou

nd?
Physics
1 answer:
ELEN [110]3 years ago
5 0
This problem can be solved using a kinematic equation. For this case, the following equation is useful:

v_final = v_initial + at

where,
v_final = final velocity of the nail
v_initial = initial velocity of the nail
a = acceleration due to gravity = 9.8 m/s^2
t = time 

First, we determine the time it takes for the nail to hit the ground. We know that the initial velocity is 0 m/s since the nail was only dropped. It has a final velocity of 26 m/s. We substitute these values to the equation and solve for t:

26 = 0 + 9.8*t
t = 26/9.8 = 2.6531 s

The problem asks the velocity of the nail at t = 1 second. We then subtract 1 second from the total time 2.6531 with v_final as unknown.

v_final = 0 + 9.8(2.6531-1) = 16.2004 m/s.

Thus, the nail was traveling at a speed of 16. 2004 m/s, 1 second before it hit the ground. 

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A tennis ball is hit into the air with a racket. When is the balls kinetic energy the greatest?
madam [21]

Answer:

Kinetic energy is maximum when the player hits the ball.

Explanation:

Kinetic energy =\frac{1}{2} mv^2, where m is the mass and v is the velocity.

So kinetic energy is proportional to square of velocity.

Velocity is maximum when the player hits the ball.

So kinetic energy is maximum when the player hits the ball.

3 0
2 years ago
We have three identical metallic spheres A, B, C. Initially sphere A is charged with charge Q, while B and C are neutral. First,
larisa [96]

Answer:

The final charges of each sphere are:   q_A = 3/8 Q , q_B = 3/8 Q ,               q_C = 3/4 Q

Explanation:

This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.

Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point

                q_A = Q / 2

                q_B = Q / 2

Now sphere A touches sphere C, ending with half the charge

                q_A = ½ (Q / 2) = ¼ Q

                q_B = ¼ Q

Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge

                  q = Q / 4 + Q / 2 = ¾ Q

This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q

                  q_A = 3/8 Q

                  q_B = 3/8 Q

The final charges of each sphere are:

                q_A = 3/8 Q

                q_B = 3/8 Q

                q_C = 3/4 Q

7 0
3 years ago
How much work is done on 10.0 c of charge to move it through a potential diffrence of 9.0 v in 10.0s
GenaCL600 [577]
We need more evidence to be provided
4 0
2 years ago
Make a rough estimate of the number of quanta emitted in one second by a 100 W light bulb. Assume that the typical wavelength em
mixas84 [53]

Answer:

#_photon = 5 10²⁰ photons / s

Explanation:

For this exercise let's calculate the energy of a single quantum of energy, use Planck's law

         E = h f

         c= λ f

         E = h c / λ

          λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m

Let's calculate

          E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹

          E₀ = 19.89 10⁻²⁰ J

This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w

                #_photon = P / E₀

               #_photon = 100 / 19.89 10⁻²⁰

              #_photon = 5 10²⁰ photons / s

6 0
3 years ago
A person is diving in a lake in the depth of h = 5.5 m. The density of the water is rho = 1.0 x10^3 kg/m^3. The pressure of the
Cloud [144]

Answer:a) P = Po + rho×h×g

b) P = 5.4 × 10^9 pa

c) F = P/A = (Po + rho×h×g)/A

d) 1.174×10^11N

Explanation: Using the formula

P = Po + rho×h×g

P =  1.0 x 10^5 + 1000 × 5.5 × 9.81

P = 5.4 × 10^9pa

The magnitude of the force exerted by water on the top of the person's head F at the depth h in terms of P

F = P/A = (Po + rho×h×g)/A

Using the above formula

Where A = 0.046m^2

F = P/ A = 5.4×10^9/0.046

F = 1.174×10^11N

3 0
3 years ago
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