All categories

3 years ago

What are the energy transfers in a torch?

Physics

1 answer:

lapo4ka [179]3 years ago

6 0

Explanation:

The battery is a store of internal energy (shown as chemical energy). The energy is transferred through the wires to the lamp, which then transfers the energy to the surroundings as light. These are the useful energy transfers - we use electric lamps to light up our rooms.

You might be interested in

Please answer these two questions ASAP!! Will mark brainliest and give points!

maw [93]

Counter clockwise torque is 360Nm.
Clockwise torque is 240Nm.

40 * 9 = 360
80 * 3 = 240

6 0

3 years ago

Read 2 more answers

The process by which you group things based on their similarities is known as classifying.

dusya [7]

True. Classifying by similarities is the basis for biological classification.

8 0

3 years ago

Give you point but please i need help in physcis

Anastasy [175]

Explanation:

BRAINILIEST PLEASE

7 0

3 years ago

Most earthquakes occur in areas close to where tectonic plates meet. There are earthquakes in San Francisco. What can be conclud

earnstyle [38]

B. To have an earthquake there must be a fault line (where two or more tectonic plates meet) so if San Fran. has earthquakes they’re on a fault line.

5 0

2 years ago

Read 2 more answers

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

3 0

3 years ago