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Masja [62]
2 years ago
9

Sandy is on a road trip. She leaves at 8:00 AM. It takes her 2 hours to drive 200 kilometers. She stops at a rest stop for half

an hour. Then, she drives for 100 more kilometers, which takes her an hour and a half. What is Sandy's average velocity?
Physics
1 answer:
lutik1710 [3]2 years ago
0 0
The average velocity of Sandy is given by the total distance covered S divided by the total time taken t:
v= \frac{S}{t}

The total distance covered is
S=200 km+0+100 km=300 km
while the total time taken is 2 hours + half an hour (for the rest) + 1 hour and half, so
t=2h+0.5h+1.5 h=4 h
Therefore, the average velocity is 
v= \frac{S}{t}= \frac{300 km}{4 h}=75 km/h
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choli [55]
Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.

Rate of speed that toy is moving is irelevant.

childs force is:
Fc = 2N
Fc = Ff  (Ff -friction force)

Ff = a*Q

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if we express a we get
a = F/Q = 2/8 = 0.25
8 0
2 years ago
Read 2 more answers
) a charge of 6.15 mc is placed at each corner of a square 0.100 m on a side. determine the magnitude and direction of the force
Nana76 [90]
Because charges are positioned on a square the force acting on one charge is the same as the force acting on all others. 
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We will break down forces on their x and y components:
F_x=F_3+F_2cos(45^{\circ})
F_y=F_1+F_2sin(45^{\circ})
Let's figure out each component:
F_1=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\
F_3=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\
F_2=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}
Total force acting on the charge would be:
F=\sqrt{F_x^2+F_y^2}
We need to calculate forces along x and y axis first( I will assume you meant micro coulombs, because otherwise we get forces that are huge).
F_x=F_3+F_2cos(45^{\circ})=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}+\frac{1} {4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot\cos(45)=46N
F_y=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot sin(45)+\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}=46N
Now we can find the total force acting on a single charge:
F=\sqrt{F_x^2+F_y^2}=\sqrt{46^2+46^2}=65N
As said before, intensity of the force acting on charges is the same for all of them.

5 0
2 years ago
A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir
ikadub [295]
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

   KE1  = KE2

The kinetic energy of the system before the collision is solved below.

  KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
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This value should also be equal to KE2, which can be calculated using the conditions after the collision.

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The value of x from the equation is 17.16 cm/s.

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6 0
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Which of the following states that all matter tends to "warp" space in its vicinity and that objects react to this warping by ch
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There are no choices on the list you provided that make such a statement,
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4 0
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Volgvan

Answer:

The gravitational field at any point is independent of the objects placed at that point such as a stone or paper.

And, also the gravitational force acting on a body is determined by the intensity of the field at that point.

Explanation:

Given,

The gravitational force between two bodies separated by a distance r is given by the relation,

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Where,

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                     r - the distance between the center of masses

Let m₁ be the mass of rock. Then force acting on the rock is given by

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The gravitational force between the stone and Earth is given by

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Equating these two forces

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Canceling out the mass of the stone.

This shows that the force component acceleration is independent of the mass of the stone.

Hence, it implies that the gravitational field at any point is independent of the objects placed at that point such as a stone or paper.

And, also the gravitational force acting on a body is determined by the intensity of the field at that point.

4 0
2 years ago
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