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mixer [17]
3 years ago
7

A student walks 31 m east, then 16 m west. Make east positive. What is their displacement?

Physics
1 answer:
Marianna [84]3 years ago
3 0
Displacement = 31 - 16 = +15 m
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Michael is looking for examples of energy at the park. He notices a mother pushing her child on a swing. He graphs the number of
inn [45]

Answer:

D.

Explanation:

As you see, the graph INCREASES, which means it goes up. It is not going down, only rising.

If I'm wrong forgive me ;-;

6 0
2 years ago
What is the direction of the force of an object moving in a circle?
Ivanshal [37]

Answer:

The answer is 45 degree angle

7 0
3 years ago
What is the formula to find the second charge in Coulomb's Law?
kicyunya [14]

Answer:

1.F is the electrostatic force between charges (in Newtons),

2.q₁ is the magnitude of the first charge (in Coulombs),

3.q₂ is the magnitude of the second charge (in Coulombs),

4.r is the shortest distance between the charges (in m),

5.ke is the Coulomb's constant. It is equal to 8.98755 × 10⁹ N·m²/C² .

5 0
2 years ago
Read 2 more answers
Charges of 4.0 μC and −6.0 μC are placed at two corners of an equilateral triangle with sides of 0.10 m. What is the magnitude o
jek_recluse [69]

Answer:

4.763 × 10⁶ N/C

Explanation:

Let E₁ be the electric field due to the 4.0 μC charge and E₂ be the electric field due to the -6.0 μC charge. At the third corner, E₁ points in the negative x direction and E₂ acts at an angle of 60 to the negative x - direction.

Resolving E₂ into horizontal and vertical components, we have

E₂cos60 as horizontal component and E₂sin60 as vertical component. E₁ has only horizontal component.

Summing the horizontal components we have

E₃ = -E₁ + (-E₂cos60) = -kq₁/r²- kq₂cos60/r²

= -k/r²(q₁ + q₂cos60)

= -k/r²(4 μC + (-6.0 μC)(1/2))

= -k/r²(4 μC - 3.0 μC)

= -k/r²(1 μC)

= -9 × 10⁹ Nm²/C²(1.0 × 10⁻⁶)/(0.10 m)²

=  -9 × 10⁵ N/C

Summing the vertical components, we have

E₄ = 0 + (-E₂sin60)

= -E₂sin60

= -kq₂sin60/r²

= -k(-6.0 μC)(0.8660)/(0.10 m)²

= -9 × 10⁹ Nm²/C²(-6.0 × 10⁻⁶)(0.8660)/(0.10 m)²

= 46.77 × 10⁵ N/C

The magnitude of the resultant electric field, E is thus

E = √(E₃² + E₄²) = √[(-9 × 10⁵ N/C)² + (46.77 10⁵ N/C)²) = (√226843.29) × 10⁴

= 476.28  × 10⁴ N/C

= 4.7628 × 10⁶ N/C

≅ 4.763 × 10⁶ N/C

8 0
3 years ago
Which situation describes the highest rate of power?
Darya [45]

Answer : The correct option is, (D) A machine does 400 joules of work in 5 seconds.

Explanation :

Power : It is defined a the rate of doing work per unit time.

Formula used :

P=\frac{w}{t}

where,

P = power

w = work done

t = time

Now we have to determine the rate of power for the following options.

(A) A machine does 200 joules of work in 10 seconds.

P=\frac{200}{10}=20W

(B) A machine does 400 joules of work in 10 seconds.

 P=\frac{400}{10}=40W

(C) A machine does 200 joules of work in 5 seconds.

 P=\frac{200}{5}=40W

(D) A machine does 400 joules of work in 5 seconds.

P=\frac{400}{5}=80W

From this we conclude that, a machine does 400 joules of work in 5 seconds has the highest rate of power.

Hence, the correct option is, (D)

8 0
3 years ago
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