All categories

3 years ago

A 43 kg box is being pushed and pulled by

Physics

1 answer:

m_a_m_a [10]3 years ago

4 0

Answer:

a = - 0.209 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

We will take the positive forces to the right and the negative forces to the left.

$155-277+113=43*a\\-9=43*a\\a = - 0.209[m/s^{2} ]$

The negative sign means that the box accelerates in a negative direction to the left.

You might be interested in

All household circuits are wired in parallel. A 1140-W toaster, a 270-W blender, and a 80-W lamp are plugged into the same outle

VARVARA [1.3K]

Answer:

total current = 12.417 A

so it will not fuse as current is less than 15 A

Explanation:

given data

toaster = 1140-W

blender = 270-W

lamp = 80-W

voltage = 120 V

solution

we know that current is express as

current = power ÷ voltage ......................1

here voltage is same in all three device

current by toaster is

I = $\frac{1140}{120}$

I = 9.5 A

and

current by blender

I = $\frac{270}{120}$

I = 2.25 A

and

current by lamp is

I = $\frac{80}{120}$

I = 0.667 A

so here device in parallel so

total current is = 9.5 A + 2.25 A + 0.667 A

total current = 12.417 A

so it will not fuse as current is less than 15 A

8 0

2 years ago

Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o

zhannawk [14.2K]

Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

We need to calculate the magnitude of charge

Using newton's second law

$F= ma$

$a=\dfrac{F}{m}$

Put the value of F

$a=\dfrac{kq^2}{mr^2}$

Put the value into the formula

$414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}$

$q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}$

$q^2=1.84\times10^{-14}$

$q=0.135\times10^{-6}\ C$

$q=0.135\ \mu C$

Hence, The magnitude of the charge on each sphere is 0.135μC.

7 0

3 years ago

Read 2 more answers

A tennis player smashes a ball of mass m horizontally at a vertical wall. The ball rebounds at the same speed v with which it st

Morgarella [4.7K]

Answer:

B.0

Explanation:

Change in momentum: This is defined as the product of mass and change in velocity of a body. or it can be defined as the product of force and time of a body. The fundamental unit of change in momentum is kg.m/s

Change in momentum = M(V-U)......................... Equation 1

where M = mass of the ball, V = final velocity of the ball, U = initial velocity of the ball.

Let: M = m kg and V = U = v m/s

Substituting these values into equation 1

Change in momentum = m(v-v)

Change in momentum = m(0)

Change in momentum = 0 kg.m/s

Therefore the momentum of the ball has not changed.

The right option is B.0

5 0

3 years ago

8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h

pantera1 [17]

(8) A car starting with a speed v skids to a stop over a distance d, which means the brakes apply an acceleration a such that

0² - v² = 2 a d → a = - v² / (2d)

Then the car comes to rest over a distance of

d = - v² / (2a)

Doubling the starting speed gives

- (2v)² / (2a) = - 4v² / (2a) = 4d

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration a such that

0² - (13.9 m/s)² = 2 a (15 m) → a ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance d such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) d → d ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass m such that

60 J = m g h

where g = 10 m/s² and h is the height it is lifted, 1.2 m. Solving for m gives

m = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

4 0

3 years ago

The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kn/m. determine the maximum shear s

alex41 [277]

Given:

Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.

We need to determine the maximum shear stress developed in the beam:

τ = F/A

Assuming the area of the beam is 100 m^2 with a length of 10 m.

τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2

8 0

3 years ago