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2 years ago

14

5- A 2500g object is pushed with 55N for 12m in 11s, there was a force of friction of 30N.

1 answer:

Assoli18 [71]2 years ago

3 0

Answer:

1kg =1000g

2.5kg

D=12m

t=11s

F=2.5KG

Explanation:

work done =f.d

=2.5×12

=30Nm

55-30

average speed

final - initial

divide by time t(s)

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umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

3 years ago

How can energy conservation be considered an energy source?

Brut [27]

Answer:This could be in the form of using fewer energy services or using devices that require less energy.

Explanation:

7 0

3 years ago

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

3 years ago

The mass of the skier including his equipment is 75 kg in the ski race, the total vertical change in height is 880m

Bogdan [553]

75 percent off of water and please water the light water and water water and then go back and please water pollution please 880m

8 0

3 years ago

Consider a venturi with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir.

Marina CMI [18]

Answer:

( $P_1-P_2$ )=1913.31 N/m^2

Explanation:

given:

$\frac{A_t}{A_1}$ =0.85

$V_1$ =90 m/s

γ∞=1.23 kg/m^3

solution:

since outside pressure is atm pressure vaccum can be defined by ( $P_1-P_2$ )

$V_1$ =√2( $P_1-P_2$ )/γ∞[ $\frac{A_t}{A_1}^2$ -1]

( $P_1-P_2$ )=1913.31 N/m^2

6 0

3 years ago