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2 years ago

Please help ASAP! Thank you :)

Physics

1 answer:

puteri [66]2 years ago

8 0

Answer:

magnitude of gravitational force between the Earth and the Sun at B is greater than that at A

Explanation:

Formula of gravitational force:

F = GMm/r^2

(r is the distance between 2 objects)

We see that r(B) < r(A) since at B, the Earth is closer to the Sun than at A

According to the Formula, the smaller r is, the greater F is

So, F(B) > F(A)

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Sound waves are also called compression waves. This means that as the wave travels through air, the ____ in creases and decrease

Alja [10]

The speed creases and decreases

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3 years ago

A massless string connects a 10.00 kg mass to a 13.00 kg cart which is resting on a frictionless horizontal surface. The mass ha

ch4aika [34]

The cart's acceleration to the right after the mass is released is determined as 7.54 m/s².

<h3>Acceleration of the cart</h3>

The acceleration of the cart is determined from the net force acting on the mass-cart system.

Upward force = Downward force

ma = mg

13a = 10(9.8)

13a = 98

a = 98/13

a = 7.54 m/s²

Thus, the cart's acceleration to the right after the mass is released is determined as 7.54 m/s².

Learn more about acceleration here: brainly.com/question/14344386

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6 0

2 years ago

the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i

Naily [24]

Answer:

(a) $\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b) P = 0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

$P = \sigma AT^4$

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

$\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b)

Now, for 90% radiator blackbody at 2000 K:

$P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4$

7 0

3 years ago

"There are two types of error that can occur when making measurements: systematic and random error. How you correct the error de

Murljashka [212]

Answer:

Systematic error can be corrected using calibration of the measurement instrument, while random error can be corrected using an average measurement from a set of measurements.

Explanation:

Random errors lead to fluctuations around the true value as a result of difficulty taking measurements, whereas systematic errors lead to predictable and consistent departures from the true value due to problems with the calibration of your equipment.

Systematic error can be corrected, by calibration of the measurement instrument. Calibration is simply a procedure where the result of measurement recorded by an instrument is compared with the measurement result of a standard value.

Random error can be corrected using an average measurement from a set of measurements or by Increasing sample size.

8 0

3 years ago

Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Sim

Nonamiya [84]

Answer:

The distance covered by puck A before collision is $z = 8.56 \ m$