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3 years ago

5

A construction worker pushes a wheelbarrow with a total mass of 50.0kg. What is the acceleration of the wheelbarrow if the net f

orce on it is 75 N West?

1 answer:

Zarrin [17]3 years ago

5 0

Did you try googling it lol thats what i do if its a problem like that. sometimes there are websites that answer it you just have to look really hard

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Answer this please.____

ad-work [718]

Answer:

Explanation:

When a liquid is heated, the particles are given more energy. They start to move faster and further apart. At a certain temperature, the particles break free of one another and the liquid turns to gas. This is the boiling point.

5 0

3 years ago

A 1200-kg SUV is moving alone a straight highway at 12.0 m/s. Another car, with mass 1800 kg and speed 20.0 m/s, has its center

andreev551 [17]

Answer:

A) d = 24 m

B) 50400 kg.m/s

C) v₀ = 16.8 m/s

D) 50400 kg.m/s. It's equal to the momentum found in part B.

Explanation:

We are given;

Mass of station wagon;m1 = 1200 kg Velocity; V = 12 m/s

Mass of car; m2 = 1800 kg

Velocity of car; v2 = 20 m/s

a ) Let centre of mass of car and station wagon be at a distance d from wagon

Thus,

If we take moment of weight about it, we have;

1200 x d = 1800 x ( 40 - d )

Where, d is the position of the center of mass of the system consisting of the two cars

Thus,

1200d = 72000 - 1800d

1200d + 1800d = 72000

3000 d = 72,000

d = 72,000/3000

d = 24 m

b ) Total momentum= m1•v1 + m2•v2

= (1200 x 12) + (1800 x 20)

= 14400 + 36000

= 50400 kg.m/s

c ) Let speed of centre of mass be v₀

Thus,

v₀ = (m1•v1 + m2•v2)/(m1 + m2)

v₀ = 50400/(1200 + 1800)

v₀ = 50400/3000

v₀ = 16.8 m/s

d) System Total momentum = velocity of centre mass x total mass

Thus,

Total momentum = v₀(m1 + m2)

= 16.8(3000) = 50400 kg.m/s .

This value is equal to what was calculated in part b

8 0

3 years ago

Consider the space between a point charge and the surface of a neutral spherical conducting shell. If the charge sits at the cen

Furkat [3]

Answer:

True

Explanation:

If a thin, spherical, conducting shell carries a negative charge, We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. The electric field-lines produced outside such a charge distribution point towards the surface of the conductor, and end on the excess electrons. Moreover, the field-lines are normal to the surface of the conductor. This must be the case, otherwise the electric field would have a component parallel to the conducting surface. Since the excess electrons are free to move through the conductor, any parallel component of the field would cause a redistribution of the charges on the shell. This process will only cease when the parallel component has been reduced to zero over the whole surface of the shell

According to Gauss law

∅ = EA =-Q/∈₀

Where ∅ is the electric flux through the gaussian surface and E is the electric field strength

If the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero. In fact, the electric field inside any closed hollow conductor is zero

8 0

3 years ago

As the wavelength increases, the frequency (2 points) decreases and energy decreases. increases and energy increases. decreases

frozen [14]

Bohr's equation for the change in energy is
$\Delta E= \frac{hc}{\lambda}$
where
h = Planck's constant
c == the velocity of light
λ = wavelength.

The velocity is related to wavelength and frequency, f, by
c = fλ

Let us examine the given answers on the basis of the given equations.

a. As λ increases, f decreases and ΔE decreases.
TRUE

b. As λ increases, f increases and ΔE increases.
FALSE

c. As λ increases, f increases and ΔE decreases.
FALSE

Answer:
As the wavelength increases, the frequency decreases and energy decreases.

3 0

3 years ago

Read 2 more answers

Consider a single-slit diffraction pattern for λ=589nm , projected on a screen that is 1.00 m from a slit of width 0.25 mm. How

const2013 [10]

Answer: i) 2.356 × 10^-3 m = 2.356mm, ii) 4.712 × 10^-3 m = 4.712mm

Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as

y = R×(mλ/d)

Where y = distance between nth fringes and the center fringe.

m = order of fringe

λ = wavelength of light = 589nm = 589×10^-9m

R = distance between slits and screen = 1.0m

d = distance between slits = 0.25mm = 0.00025m

For distance between the first dark fringe and the center fringe.

This implies that m = 1

y = 1 × 589×10^-9 × 1/0.00025

y = 589×10^-9/0.00025

y = 2,356,000 × 10^-9

y = 2.356 × 10^-3 m = 2.356mm

For the second dark fringe, this implies that m = 2

y = 1 × 2 × 589×10^-9/0.00025

y = 1178 × 10^-9 /0.00025

y = 4,712,000 × 10^-9

y = 4.712 × 10^-3 m = 4.712mm

8 0

3 years ago