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andreev551 [17]
3 years ago
9

Alexandra is dragging her 37.8 kg golden retriever cross the wooden floor by applying a horizontal force. What force must be app

lied to move a dog with a constant speed of 1 m/s? The coefficient of kinetic friction between the dog in the floor is one point
Physics
1 answer:
Mila [183]3 years ago
6 0

Answer:

F = 385.56 N

Explanation:

Given that,

Mass, m = 37.8 kg

He applies a horizontal force and cross the wooden floor.

We need to find the force that must be applied to move a dog with a constant speed of 1 m/s.

The coefficient of kinetic friction between the dog in the floor is 1.02.

The net force acting on it is given by :

F = f- μmg

f is force due to constant speed, f = 0 (since, a = 0)

F = μmg

= 1.02 × 37.8 × 10

= 385.56 N

Hence, the required force is 385.56 N.

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A rope vibrates every 0.5 s. what is the frequency of the waves?
umka2103 [35]
F = 1/t
F = 1/0.5
F = 2Hz

Answer is 2Hz
4 0
3 years ago
Read 2 more answers
Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of
Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = \frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = \sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}

(|r₂₁|)² = 148.25

F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }

      = 0.050938(0.19107i + 0.54933j) N

      = (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0
3 years ago
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
3 years ago
How are energy time and power related physics?
olganol [36]
Power is the rate of energy. Mathematically, it is

Power (p) = Energy(E) / Time(t)

Hope this helps!
6 0
3 years ago
At which point on the image to the right would the ball have the greatest velocity if it moved from A to G.
marusya05 [52]

Answer:

It would be A, because it is has more height in which the potential energy would be greater.

4 0
3 years ago
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