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Monica [59]
3 years ago
11

amping equipment weighing 6.0 kN is pulled across a frozen lake by means of ahorizontal rope. The coecient of kinetic friction i

s 0.05. How much work is done by thecampers in pulling the equipment 1.00 km if its speed is increasing at the constant rate of0.20m/s2?
Physics
1 answer:
Harman [31]3 years ago
4 0

Answer:

Work done = 422.45 kJ

Explanation:

given,                                  

weight of equipment = 6 kN      

coefficient of kinetic friction = 0.05          

distance up to which it is pulled = 1000 m

constant acceleration = 0.2 m/s²                    

Work done by the camper = ?                

actual acceleration acting a'      

m a = m a' - μ mg            

a' = a + μ g                      

a' = 0.2 + 0.05 x 9.8                

a' = 0.69 m/s²                              

Work done = Force x distance

F = m a'                                                            

F = \dfrac{6000}{9.8} \times 0.69

F = 422.44897 N                            

Work done = F x d                          

Work done = 422.44897 x 1000

Work done = 422449 J                  

Work done = 422.45 kJ            

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mars1129 [50]

Answer:

Answer:

New speed of the 22-kg block is 1.57 m/s

Explanation:

Mass of block  

Mass of another block  

Initial speed of the block  

Initial speed  of the another block  

Initial speed  of the another block  

For conservation of momentum, we have

Substitute all the values and solving for final speed of the 22kg block is

new speed of the 22-kg block is 1.57 m/s

Couldnt write the answer so check picture

8 0
3 years ago
6. A car is traveling at 50m/s when it begins to slow down to
ki77a [65]

Answer:

-6 m/s²

Explanation:

Given:

v₀ = 50 m/s

v = 20 m/s

t = 5 s

Find: a

v = at + v₀

20 m/s = a (5 s) + 50 m/s

a = -6 m/s²

8 0
3 years ago
A spherical hot air balloon of diameter 30.0 meters just floats (hovers) when the hot air inside has been heated to a density of
Rom4ik [11]

Answer:

W = 166422.729 N

Explanation:

given,

diameter of the balloon = 30 m

density of the air = 1.10 Kg/m³

weight of the balloon and cargo = ?

density of the surrounding air = 1.20 kg/m³

we know,

Density = mass/volume

m = density x volume

m = \rho\times \dfrac{4}{3}\pi r^3

m =1.20 \times \dfrac{4}{3}\pi\times 15^3

m = 16964.6 Kg

Weight of the balloon

 W = m g

 W = 16964.6 x 9.81

W = 166422.729 N

Weight of the balloon and the cargo is equal to  W = 166422.729 N

5 0
3 years ago
A length of copper wire carries a current of 11 A, uniformly distributed through its cross section. Calculate the energy density
NISA [10]

Answer:

a)0.983 \frac{J}{m^3}

b)u_E =7.329x10^-3 \frac{J}{m^3}

Explanation:

The energy density is "the energy per unit volume, in the electric field.  The energy stored between the plates of the capacitor equals the energy per unit volume stored in the electric field times the volume between the plates".

A magnetic field is a "vector field that describes the magnetic influence of electric charges in relative motion and magnetized materials".

Part a

For this case we can assume use the equation for the magnetic field in terms of the energy per unit of volume.

B=\sqrt{2\mu_o u}

Where μ0 represent the permeability constant, also known as the magnetic constant. If we solve for u we got:

u=\frac{B^2}{2\mu_o}

We also know that the magnetic field can be expressed in terms of the current and the radius of action R like this:

B=\frac{\mu_o i}{2\pi R}

Replacing this on the formula for u we have:

u=\frac{1}{2\mu_o}(\frac{\mu_o i}{2\pi R})^2

And simplyfing we got:

u=\frac{\mu_o i^2}{8\pi^2 R^2}

Replacing the values given we have:

u=\frac{(4\pix10^{-7} \frac{H}{m} (11A)^2}{8\pi^2 (0.0014m)^2} =0.983 \frac{J}{m^3}

Part b

The density current is given by this formula J=i/A and the resistance by R=\frac{\rho l}{A}

If we use the equation for the energy density we have this:

u_E =\frac{1}{2}\varepsilon_o E^2 =\frac{\varepsilon}{2}(\rho J)^2=\frac{\varepsilon}{2}(\frac{iR}{l})^2

And replacing the values given we have:

u_E =\frac{8.85x10^{-12}\frac{F}{m}}{2}(\frac{11A(3700\frac{\Omega}{m})}{l})^2 =7.329x10^-3 \frac{J}{m^3}

4 0
3 years ago
A small fish is dropped by a pelican that is rising steadily at 0.500 m/s. How far below the pelican is the fish after 2.50 s?
sesenic [268]
Refer to the diagram shown below.

h = original height of the pelican when the fish is dropped (not relevant).
S =  distance traveled by the fish as a function of time, measured upward.
u = 0.5 m/s, the upward velocity with which the fish is dropped.
g = 9.8 m/s², the acceleration due to gravity.

Use the following equation:
S = ut + (1/2)gt²

S = (0.5 m/s)*(2.5 s) + 0.5*(-9.8 m/s²)*(2.5 s)²
   = -29.375 m

The negative sign means that the fish drops by  29.375 m from the original height of h.

Answer: The fish is 29.375 m below where the pelican dropped it after 2.5 s.

7 0
3 years ago
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