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iragen [17]
3 years ago
13

You are on earth, and have an object with a mass of 7 kg. with how much force does the earth's gravity pull the object downward?

Physics
1 answer:
svp [43]3 years ago
6 0
This is equivalent to asking the mass’s weight. Force=mass * acceleration. So 7*9.8= 68.6 N
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PLEASE HELP WILL MARK BRAINLIEST
loris [4]

Answer:

Protium

Explanation: Protium is an isotope of hydrogen that is composed of one proton and one electron. It is the most abundant form of hydrogen.

4 0
3 years ago
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A 5.0 Ω resistor is hooked up in series with a 10.0 Ω resistor followed by a 20.0 Ω resistor. The circuit is powered by a 9.0 V
yan [13]
<h2>Answer:</h2>

(a) Attached to the response as Figure 1.

(b) 35.0Ω

(c) Across 5.0Ω = 1.3V

   Across 10.0Ω = 2.6Ω

   Across 20.0Ω = 5.2Ω

<h2>Explanation:</h2>

(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.

(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e

R = R₁ + R₂ + R₃               -------------------(i)

Where;

R₁ = 5.0 Ω

R₂ = 10.0 Ω

R₃ = 20.0 Ω

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

∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω

∴ R = 35.0 Ω

Therefore, the equivalent resistance is ∴ R = 35.0Ω

(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;

i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.

ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;

V = I R              ------------------(ii)

Where;

V = Voltage supplied to the circuit = 9.0V

I = Current through the circuit

R = Resistance of the equivalent resistor = 35.0Ω

Substitute these values into equation (ii)

9.0 = I x 35.0

I = \frac{9.0}{35.0}

I = 0.26A

This is also the current flowing through each of the resistors separately.

iii. Calculate the voltage drop across

1.<em> 5.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 5.0Ω resistor

I = current through the 5.0Ω resistor = 0.26A

R = resistance of the 5.0Ω resistor = 5.0Ω

=> V = 0.26 x 5.0

=> V = 1.3V

2.<em> 10.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 10.0Ω resistor

I = current through the 10.0Ω resistor = 0.26A

R = resistance of the 10.0Ω resistor = 10.0Ω

=> V = 0.26 x 10.0

=> V = 2.6V

3.<em> 20.0 Ω resistor</em>

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 20.0Ω resistor

I = current through the 20.0Ω resistor = 0.26A

R = resistance of the 20.0Ω resistor = 10.0Ω

=> V = 0.26 x 20.0

=> V = 5.2V

7 0
3 years ago
Can someone help me on 2 science question,
gavmur [86]
1. I think you should compare diagrams of moon phases from the textbook to diagrams of moon phases online. Because if you pick D it will take to long and C will help you out whith 3 different things to look at. 

2. The moon changes in appearances from the perspective of people on earth because it's revolving around the planet and the earth is revolving around the sun, so  A. Hoped this helped.

8 0
3 years ago
Read 2 more answers
A rock is thrown from the top of a 20-m building at an angle of 53° above the horizontal. If the horizontal range of the throw
NemiM [27]

Answer:

28.5 m/s

18.22 m/s

Explanation:

h = 20 m, R = 20 m, theta = 53 degree

Let the speed of throwing is u and the speed with which it strikes the ground is v.

Horizontal distance, R = horizontal velocity x time

Let t be the time taken

20 = u Cos 53 x t

u t = 20/0.6 = 33.33 ..... (1)

Now use second equation of motion in vertical direction

h = u Sin 53 t - 1/2 g t^2

20 = 33.33 x 0.8 - 4.9 t^2     (ut = 33.33 from equation 1)

t = 1.17 s

Put in equation (1)

u = 33.33 / 1.17 = 28.5 m/s

Let v be the velocity just before striking the ground

vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s

vy = uSin 53 - 9.8 x 1.17

vy = 28.5 x 0.8 - 16.66

vy = 6.14 m/s

v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2

v = 18.22 m/s

6 0
3 years ago
HELP ASAP PLZ!
Dafna1 [17]
1. is D. Scientific method cannot be replaced. 
2. is B. 
3. is B as well.
8 0
2 years ago
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