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mylen [45]
3 years ago
11

How do you calculate joules? someone please help me I beg

Physics
1 answer:
Gnoma [55]3 years ago
5 0

Answer:

<h2><em><u>Joules</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>Watts</u></em><em><u> </u></em><em><u>×</u></em><em><u> </u></em><em><u>Seconds</u></em></h2>

Explanation:

If you multiply the number of watts by the number of seconds, you'll end up with joules. 

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Even at such low fractions of the speed of light these stars are moving quite quickly (thousands of kilometers a second) compare
yulyashka [42]
According to Doppler Effect, an observer at rest will perceive a shift in the wavelength or frequency of the radiation emitted by a source in movement.This shift is given by the formula:
\frac{ \lambda - \lambda_{0} }{ \lambda_{0} } = \frac{-v}{c}

where:
\lambda = observed wavelength
\lambda_{0} = wavelength at rest
v = speed of source (positive if towards the observer, negative if away from the observer)
c = speed of light

Therefore, we can solve for the observed wavelength:
\lambda = \lambda_{0} (\frac{-v}{c}) + \lambda_{0} \\ \lambda = \lambda_{0} (1 - \frac{v}{c})

Substituting the given data:
\lambda = 656.46 (1 - \frac{300}{299792})
= 655.80 nm

Hence, the observed wavelength of the line would be 655.80 nm. Note that this value is smaller than the one at rest, which means that we have a blue-shift, as expected for an approaching source.
5 0
3 years ago
The formula for electrical power is p = i v. Which is the equivalent equation solved for i? startfraction p over v endfraction =
Anettt [7]

The equivalent formula for p = iv is the startfraction p over v endfraction = i

In physics, power measures the rate of transfer of electrical energy through a circuit per unit of time. It is denoted by P and is measured using the SI unit of power (watts or 1 joule/second). Electrical energy is usually supplied by batteries and produced by generators.

There are three types of power, true, reactive, and apparent, which are related in trigonometric form. This is called the power triangle. A power triangle that relates apparent power to active and reactive power. Power is measured in watts. One watt equals one joule per second.

Learn more about the power in

brainly.com/question/24110214

#SPJ4

8 0
1 year ago
The spring constant, k, for a 22cm spring is 50N/m. A force is used to stretch the spring and when it is measured again it is 32
lutik1710 [3]

Answer:

5N

Explanation:

Given parameters:

Original length = 22cm

Spring constant, K  = 50N/m

New length = 32cm

Unknown

Force applied  = ?

Solution:

The force applied on a spring can be derived using the expression below;

   Force  = KE

 k is the spring constant

 E is the extension

  extension = new length - original length

  extension  = 32cm  - 22cm  = 10cm

convert the extension from cm to m;  

   100cm  = 1m;

    10cm will give 0.1m

So;

  Force  = 50N/m x 0.1m  = 5N

8 0
3 years ago
An object is formed by attaching a uniform, thin rod with a mass of mr = 6.85 kg and length L = 5.76 m to a uniform sphere with
Ket [755]

Answer:

Part a)

I = 1879.7 kg m^2

Part b)

\alpha = 0.70 rad/s^2

Part c)

I = 153.8 kg m^2

Part 4)

angular acceleration will be ZERO

Part 5)

I = 345.6 kg m^2

Explanation:

Part a)

Moment of inertia of the system about left end of the rod is given as

I = \frac{m_r L^2}{3} + (\frac{2}{5} m_s R^2 + m_s(R + L)^2)

So we have

I = \frac{m_r(4R)^2}{3} + (\frac{2}{5}(5m_r) R^2 + (5m_r)(R + 4R)^2)

I = \frac{16}{3}m_r R^2 + (2m_r R^2 + 125 m_rR^2)

I = (\frac{16}{3} + 127)m_r R^2

I = (\frac{16}{3} + 127)(6.85)(1.44)^2

I = 1879.7 kg m^2

Part b)

If force is applied to the mid point of the rod

so the torque on the rod is given as

\tau = F\frac{L}{2}

\tau = 460(2R)

\tau = 460 \times 2 \times 1.44

\tau = 1324.8 Nm

now angular acceleration is given as

\alpha = \frac{\tau}{I}

\alpha = \frac{1324.8}{1879.7}

\alpha = 0.70 rad/s^2

Part c)

position of center of mass of rod and sphere is given from the center of the sphere as

x = \frac{m_r}{m_r + m_s}(\frac{L}{2} + R)

x = \frac{m_r}{6 m_r}(3R) = \frac{R}{2}

so moment of inertia about this position is given as

I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + \frac{R}{2})^2 + (\frac{2}{5} m_s R^2 + m_s(\frac{R}{2})^2)

so we have

I = \frac{m_r (16R^2)}{12} + m_r(\frac{5R}{2})^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(\frac{R^2}{4})

I = m_r R^2(\frac{16}{12} + \frac{25}{4} + 2 + \frac{5}{4})

I = 6.85(1.44)^2\times 10.83

I = 153.8 kg m^2

Part 4)

If force is applied parallel to the length of rod

then we have

\tau = \vec r \times \vec F

\tau = 0

so angular acceleration will be ZERO

Part 5)

moment of inertia about right edge of the sphere is given as

I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + 2R)^2 + (\frac{2}{5} m_s R^2 + m_s(R)^2)

so we have

I = \frac{m_r (16R^2)}{12} + m_r(4R)^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(R^2)

I = m_r R^2(\frac{16}{12} + 16 + 2 + 5)

I = 6.85(1.44)^2\times 24.33

I = 345.6 kg m^2

6 0
3 years ago
A vehicle ___________ occurs when the driver begins to back up and is unaware that a person is behind the vehicle.
jeka94
I don't know accident?
6 0
4 years ago
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