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2 years ago

Find the weight of a 15 kg rock. Acceleration due to gravity is 9.8 m/s2.

Physics

2 answers:

Mice21 [21]2 years ago

8 0

Weight = Mass × Gravity

Mass = 15 kg

Gravity = 9.8 m/s^2

Weight = 15 kg × 9.81 m/s^2 = 147 N

Andrews [41]2 years ago

3 0

Your answer would be C. 147 N

15 kg × 9.81 m/s^2 = 147 N

Give SonOfZeus Brainliest

You deserve it for your awesome explanation! :)

(I can confirm this is correct I took the test and this was the answer :)

Have a Nice Day! :) ♥

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The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c

Flauer [41]

Answer:

$EMF = 33880 Volts$

Explanation:

As per Faraday's law of Electromagnetic induction we know that

Rate of change in magnetic flux will induce EMF in the closed conducting loop

so we have

$\phi = B.A$

now we have

$A = (110 \times 10^3)(110 \times 10^3)$

$A = 1.21 \times 10^{10}$

now we have

$\phi = B(1.21 \times 10^{10})$

now the induced EMF through this loop is given as

$EMF = (\frac{dB}{dt})(1.21 \times 10^{10})$

$EMF = (2.8 \times 10^{-6})(1.21 \times 10^{10})$

$EMF = 33880 Volts$

5 0

2 years ago

If it were possible to move a star towards the earth then its apparent magnitude number would ______ while its absolute magnitud

ArbitrLikvidat [17]

If it were possible to move a star towards the earth then its apparent magnitude number would decrease while its absolute magnitude number would stay the same.

Definition of apparent magnitude:

The luminosity of a celestial body (such as a star) as observed from the earth compare absolute magnitude.

So for example, the apparent magnitude of the Sun is -26.7 and is the brightest celestial object we can see from Earth. However, if the Sun were 10 parsecs away, its apparent magnitude would be +4.7, only about as bright as Ganymede appears to us on Earth.

Definition of absolute magnitude:

Absolute magnitude is a measure of the luminosity of a celestial object on an inverse logarithmic astronomical magnitude scale.

To learn more about apparent magnitude here

brainly.com/question/2949443

#SPJ4

5 0

1 year ago

(MICROWAVE) <br><br><br>Describe the device and its function.

andreev551 [17]

Answer: You use it to cook food so you can eat. put whatever kind of food you like in there and heat it up for how long you want and then it will beep when done. then you can eat!

8 0

2 years ago

jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh

Greeley [361]

Answer:

C. 590 mph

$\vert v_{cj}\vert=589.49\ mph$

Explanation:

Given:

velocity of jet, $v_j=500\ mph$
direction of velocity of jet, east relative to the ground
velocity of Cessna, $v_c=150\ mph$
direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

$\vec v_j=500\ \hat i\ mph$

$\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)$

$\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph$

Now the vector of relative velocity of Cessna with respect to jet:

$\vec v_{cj}=\vec v_j-\vec v_c$

$\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )$

$\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph$

Now the magnitude of this velocity:

$\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}$

$\vert v_{cj}\vert=589.49\ mph$ is the relative velocity of Cessna with respect to the jet.

8 0

3 years ago

I will be so thankful if u answer correctly!!

olga_2 [115]

<h2>Answer:</h2>

(a) 10N

<h2>Explanation:</h2>

The sketch of the two cases has been attached to this response.

<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>

In this case, a frictional force $F_{r}$ is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;

∑F = ma -------------------(i)

Where;

∑F = effective force acting on the object (box)

m = mass of the object

a = acceleration of the object

∑F = F - $F_{r}$

m = 50kg

a = 0 [At constant velocity, acceleration is zero]

<em>Substitute these values into equation (i) as follows;</em>

F - $F_{r}$ = m x a

F - $F_{r}$ = 50 x 0

F - $F_{r}$ = 0

F = $F_{r}$ -------------------(ii)

<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>

In this case, the same frictional force $F_{r}$ is opposing the movement of the box.

∑F = 1.5F - $F_{r}$

m = 50kg

a = 0.1m/s²

<em>Substitute these values into equation (i) as follows;</em>

1.5F - $F_{r}$ = m x a

1.5F - $F_{r}$ = 50 x 0.1

1.5F - $F_{r}$ = 5 ---------------------(iii)

<em>Substitute </em> $F_{r}$ <em> = F from equation (ii) into equation (iii) as follows;</em>

1.5F - F = 5

0.5F = 5

F = 5 / 0.5

F = 10N

Therefore, the value of F is 10N

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4 0

2 years ago