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svet-max [94.6K]
3 years ago
13

a 6.5 kg box is pulled across a floor by a horizontal 12.0 N tension force. If there is a 5 N friction exerted on the box, what

is its acceleration
Physics
1 answer:
Fudgin [204]3 years ago
6 0

Explanation:

Fnet = 12.0N - 5.0N = 7.0N

a = Fnet/m = 7.0N/6.5kg = 1.07m/s².

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A spaceship accelerates from 0m/s to 40m/s in 5 seconds. What is the acceleration of the spaceship
nadya68 [22]
Acceleration = change in velocity/time
= 40/5
=8m/s^2
6 0
4 years ago
A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 54 m in
Lana71 [14]

Answer:

The average power is calculated as 735.0 W

Solution:

As per the question:

Total mass, M = 1200 kg

Counter mass of the elevator, m = 950

Distance traveled by the elevator, d = 54 m

Time taken, t = 3 min = 180 s

Now,

To calculate the average power:

First, we find the force needed for lifting the weight:

Force, F = (M - m)g = (1200 - 950)\times 9.8 = 2450 N

Now, the work done by this force:

W = Fd = 2450\times 54 = 132300\ J = 132.3\ kJ

Average power is given as:

P_{avg} = \frac{W}{t} = \frac{132300}{180} = 735.0\ W

4 0
3 years ago
A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (
Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants (v_{oy} = 0) zero, so let's use the equation

     y =v_{oy} t -1/2 g t²

     y= - ½ g t²

     t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

     x = vox t

     x = v₀ₓ √2y/g

c) Speeds before touching the ground

     vₓ = vox = constant

     v_{y} = v_{oy} - gt

     v_{y} = 0 - g √2y/g

    v_{y}  = - √2gy

    tan θ = Vy / vx

    θ = tan⁻¹ (vy / vx)

    θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0
4 years ago
The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the
FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

B=\frac{2\pi*a }{u*I}

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a  = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

I=\frac{2\pi*a*B }{u} Formula(2)

B=0.2*10^{-5}  T = 0.2*10^{-5} \frac{weber}{m^{2} }

a  =8cm=0.08m

u=4*\pi *10^{-7} \frac{Weber}{A*m}

We replace the known information in the formula (2)

I=\frac{2\pi*0.08*0.2*10^{-5}  }{4\pi *10x^{-7} }

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0
3 years ago
A stone is thrown horizontally from the top of an inclined plane (angle of inclination θ). How would I find the initial speed of
Katen [24]

Answer:

S = V t     where S is the horizontal distance traveled

1/2 g t^2 = H       where H is the vertical distance traveled

t^2 = 2 H / g

V^2 = S^2 / t^2 = S^2 g / (2 H) combining equations

tan theta = H / S

V^2 = S g / (2 tan theta)

Using S = L cos theta

V^2 = L g cos theta / (2 tan theta)

Giving V in terms of L and theta

7 0
2 years ago
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