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Mazyrski [523]
2 years ago
7

How much energy does a 100 W light bulb use in 24 hours?? Show all your work....

Physics
1 answer:
GalinKa [24]2 years ago
7 0
How much does it use per hour or minute or second? its impossible to answer this question i think
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A GPS signal travels at the speed of light 300,000,000 m/s. If it takes .05s to reach a phone on the surface of the Earth, how f
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3 years ago
Please help on this one?
Nonamiya [84]

30 or c because 60-30=30

3 0
3 years ago
During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.22 s, how high does it rise? The accel
BigorU [14]

Answer:

47.4 m

Explanation:

When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.

In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

t=\frac{6.22}{2}=3.11 s

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

s=ut+\frac{1}{2}gt^2

where

s is the vertical displacement

u = 0 is the initial velocity

t = 3.11 s is the time

g=9.8 m/s^2 is the acceleration of gravity (taking downward as positive direction)

Solving the  formula, we find

s=\frac{1}{2}(9.8)(3.11)^2=47.4 m

7 0
3 years ago
How much mass should be attached to a vertical ideal spring having a spring constant (force constant)of 39.5 N/m so that it will
nordsb [41]

Answer:

m = 1 kg

Explanation:

Given that,

The force constant of the spring, k = 39.5 N/m

The frequency of oscillation, f = 1 Hz

The frequency of oscillation is given by the formula as formula as follows :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f^2=\dfrac{k}{4m\pi^2}\\\\m=\dfrac{k}{4\pi^2 f^2}\\\\m=\dfrac{39.5}{4\pi^2 \times (1)^2}\\\\m=1\ kg

So, the mass that is attached to the spring is 1 kg.

6 0
2 years ago
An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA.
rewona [7]

Answer:

the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

Explanation:

Given that;

diameter D = 2.0 mm

current I = 1.0 mA

K.E of each proton is 20 MeV

the number density of the protons in the beam = ?

Now, we make use of the relation between current and drift velocity

I = MeAv ⇒ 1 / eAv

The kinetic energy of protons is given by;

K = \frac{1}{2}m_{p}v²

v = √( 2K / m_{p} )

lets relate the cross-sectional area A of the beam to its diameter D;

A = \frac{1}{4}πD²

now, we substitute for v and A

n = I / \frac{1}{4}πeD² ×√( 2K / m_{p} )

n = 4I/π eD² × √(m_{p} / 2K )

so we plug in our values;

n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )

n =  1.98695 × 10¹⁸ × 1.6157967  × 10⁻⁵

n = 3.2 × 10¹³ m⁻³  

Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³

8 0
2 years ago
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