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olga55 [171]
3 years ago
13

150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the wag

on, the coefficient of static friction between the box and the wagon's surface is 0.600, and the coefficient of kinetic friction is 0.400. The friction force on this box is closest to_________
Physics
1 answer:
mafiozo [28]3 years ago
3 0

Answer:

the friction force on this box is closest to 45.9 N.

Explanation:

given data

Weight of the box W = 150 N

accelerating uniformly = 3.00 m/s²

coefficient of kinetic friction = 0.400

coefficient of static friction = 0.600

solution

we know box does not move relative to the wagon

we get here friction force  that is express as

friction force = mass × acceleration ..............1

here mass = weight ÷ g

mass = \frac{150}{9.8} = 15.3 kg

put value in equation 1 we get

friction force = 15.3 kg × 3

friction force =  45.9 N

So, the friction force on this box is closest to 45.9 N.

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4.) What are examples of physical and chemical properties?
lubasha [3.4K]

Answer:

Some physical properties are odor (Sewage smell), and melting and boiling point (boiling water melting ice).

Some chemical properties are  flammability (burning wood), and oxidation (rotting apple).

Explanation:

7 0
3 years ago
Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th
Sergio039 [100]

Answer:

The magnitude of the electric field are 2.38\times10^{4}\ N/C and 1.09\times10^{4}\ N/C

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell q_{inn}=3.50\times10^{-8}\ C

Charge on outer shell q_{out}=1.60\times10^{-8}\ C

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

E=\dfrac{kq}{r^2}

Where, q = charge

k = constant

r = distance

Put the value into the formula

E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}

E=2.38\times10^{4}\ N/C

The total charge enclosed  by a radial distance 20.5 cm

The total charge is

q=q_{inn}+q_{out}

Put the value into the formula

q=3.50\times10^{-8}+1.60\times10^{-8}

q=5.1\times10^{-8}\ C

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}

E=1.09\times10^{4}\ N/C

Hence, The magnitude of the electric field are 2.38\times10^{4}\ N/C and 1.09\times10^{4}\ N/C

7 0
3 years ago
If a toy has a mass of 1 kg, it has an Earth weight of​
marshall27 [118]

Answer:

The answer is 9.8 N

Explanation:

As we know that the weight of an object is the amount of gravitational force acting on the object in an upward direction if the weight is acting is the downward direction.

The formula of weight:

W = Mass x Gravitational force

W = m x g

Given data:

Mass =1 kg

g = 9.8 ms-2

W = 1kg x 9.8 ms-2 = 9.8 kgms-2 ( 1 kgms-2 = N)

SO,  

W = 9.8 N

The toy has an earth weight of 9.8 N.

4 0
3 years ago
A physics book slides off a table at 1.25ms and hits the ground after 0.4s
tresset_1 [31]

Answer:

Whats the question here?

7 0
3 years ago
A 2100-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5.00 m before coming into contact w
aliya0001 [1]

Answer:

   f = 878,080 N

Explanation:

mass of pile driver (m) = 2100 kg

distance of pile driver to steel beam (s) = 5 m

depth of steel driven (d) = 12 cm = 0.12 m

acceleration due to gravity (g0 = 9.8 m/s^{2}

calculate the average force exerted on the pile driver by the beam.

  • from work done = force x distance
  • work done = change in potential energy of the pile driver
  • equating the two equations above we have

               force x distance = m x g x (s - d)

              f x 0.12 = 2100 x 9.8 x (5- (-0.12))

              d = - 0.12 because the steel beam went down at we are taking its  

              initial position to be an origin point which is 0

              f = ( 2100 x 9.8 x (5- (-0.12)) ) ÷ 0.12

                   f = 878,080 N

4 0
3 years ago
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