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Mama L [17]
3 years ago
12

Join dua-odmb-pwp meet plßsssssss​

Physics
1 answer:
Serga [27]3 years ago
8 0

Answer:

sorry- but what........?!

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How many joules of electrical energy is transferred per second by 6V 0.5 A lamp?
Degger [83]

When you ask for "joules per second", you're asking for "watts".
The rate of energy "transfer" is 'power'.  In this case, the light bulb
transfers energy out of the electrical circuit and into the space
around it, in the form of light and heat radiation.

Electrical power = (voltage) x (current) =

                              (6 volts) x (0.5 ampere) =

                                          3 watts  =  3 joules per second.
 
4 0
3 years ago
using newtons law a force of 250N is applied to an object that accelerates at a rate of 5M/s2 what is the mass of the object?
AURORKA [14]

Answer:

50 kg

Explanation:

F = ma

250 N = m (5 m/s²)

m = 50 kg

7 0
3 years ago
The massless spring of a spring gun has a force constant k=12~\text{N/cm}k=12 N/cm. When the gun is aimed vertically, a 15-g pro
ASHA 777 [7]

Answer:

0.011 m.

Explanation:

Energy stored in the spring = Energy of the projectile.

1/2ke² = mgh ................ Equation 1

Where k = spring constant, e = extension or compression, m = mass of the projectile, g = acceleration due to gravity, h = height.

make e the subject of the equation

e = √(2mgh/k)............................. Equation 2

Given: k = 12 N/cm = 1200 N/m, m = 15 g = 0.015 kg, h = 5.0 m

Constant: g = 9.8 m/s²

Substitute into equation 2

e = √(2×0.015×5/1200)

e = √(0.15/1200)

e = √(0.000125)

e = 0.011 m.

4 0
3 years ago
Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and fi
steposvetlana [31]

Answer:

W=K_f-K_i

Explanation:

The work done on a particle by external forces is defined as:

W=\int\limits^{r_f}_{r_i} {F\cdot dr} \,

According to Newton's second law F=ma. Thus:

W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\

Acceleration is defined as the derivative of the speed with respect to time:

W=m\int\limits^{r_f}_{r_i}{\frac{dv}{dt}\cdot dr} \,\\\\W=m\int\limits^{r_f}_{r_i}{dv \cdot \frac{dr}{dt}} \,

Speed is defined as the derivative of the position with respect to time:

W=m\int\limits^{v_f}_{v_i} v \cdot dv \,

Kinetic energy is defined as K=\frac{mv^2}{2}:

W=m\frac{v_f^2}{2}-m\frac{v_i^2}{2}\\W=K_f-K_i

3 0
3 years ago
How did the magnet’s density measurement using the Archimedes’ Principle compare to the density measurement using the calculated
guapka [62]

Answer:

The two methods will yield different results as one is subject to experimental errors that us the Archimedes method of measurement, the the density measurement method will be more accurate

Explanation:

This is because the density method using the calculated volume will huve room for less errors that's occur in practical method i.e Archimedes method due to human error

5 0
3 years ago
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