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Savatey [412]
3 years ago
8

A woman takes her dog Rover for a walk on a leash. To get the little pooch moving forward, she pulls on the leash with a force o

f 20.0 N at an angle of 37° above the horizontal. How much force is tending to pull Rover forward?
A. 20.0 N


B. 16.0 N


C. 42.1 N


D. 12.0 N

Physics
1 answer:
stepladder [879]3 years ago
4 0

<u>Answer:</u>

15.97 N force is tending to pull Rover forward

<u>Explanation:</u>

 The woman pulls on the leash with a force of 20.0 N at an angle of 37° above the horizontal. The arrangement is shown in the given figure,

 We nee to find the pulling force P. The 20.0 N force has two components, 20.0 cos 37 in horizontal direction and 20.0 sin 37 in vertical direction.

  The horizontal component is equal to pulling force P, which will pull Rover forward/

  So, P = 20.0 cos 37 = 15.97 N

 15.97 N force is tending to pull Rover forward.

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Statement that combines energy and mass into one law
inysia [295]

Answer:

The law of conservation of mass or principle of mass conservation

Explanation:

It states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as the system's mass cannot change, so quantity can neither be added nor be removed.

5 0
3 years ago
A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the
barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q,  is expressed as:

KE=\frac{kQq}{r}\     \ \ \ \ \ \ ...i

k is Coulomb's constant.

#The electric field,E at radius r is expressed as:

E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii

From i and ii, we have:

KE=Eqr

r=(KE)/Eq

#Substitute actual values in our equation:

r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0
3 years ago
Read 2 more answers
Sound travels through air at 343 m/s,
sasho [114]

The sound wave will have traveled 2565 m  farther in water than in air.

Answer:

Explanation:

It is known that distance covered by any object is directly proportional to the velocity of the object and the time taken to cover that distance.

Distance = Velocity × Time.

So if time is kept constant, then the distance covered by a wave can vary depending on the velocity of the wave.

As we can see in the present case, the velocity of sound wave in air is 343 m/s. So in 2.25 s, the sound wave will be able to cover the distance as shown below.

Distance = 343 × 2.25 =771.75 m

And for the sound wave travelling in fresh water, the velocity is given as 1483 m/s. So in a time interval of 2.25 s, the distance can be determined as the product of velocity and time.

Distance = 1483×2.25=3337 m.

Since, the velocity of sound wave travelling in fresh water is greater than the sound wave travelling in air, the distance traveled by sound wave in fresh water will be greater.

Difference in distance covered in water and air = 3337-772 m = 2565 m

So the sound wave will have traveled 2565 m  farther in water than in air.

5 0
3 years ago
Does anyone know? Please help! Thank you!
Vedmedyk [2.9K]

B density will increase

7 0
3 years ago
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weegy a 7.5kg block is placed on a table. if its bottom surface area is 0.6m2 , how much pressure does the block exert on the ta
Lesechka [4]

The pressure exerted by the block on the table is given by:

p=\frac{W}{A}

where W is the weight of the box, and A is the bottom surface area of the box.

The weight of the box is: W=mg=(7.5 kg)(9.81 m/s^2)=73.6 N

Substituting into the first equation, we find the pressure:

p=\frac{W}{A}=\frac{73.6 N}{0.6 m^2}=122.7 Pa

4 0
3 years ago
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