What form of energy is a bonfire and a bunsen burner?

A worker pushes a 1.50 x 10^3 N crate with a horizontal force of 345 N a distance of 24.0 m. Assume the coefficient of kinetic f

Stella [2.4K]

(a) work=Fd
345x24=8280J 

(b) work=Force of friction*d 
Force of friction =coefficient*normal force=.22x1.5x10^3=330 
330*d=7920J 

(c) net work =8280-7920=360J

5 0

2 years ago

Read 2 more answers

As the driver steps on the gas pedal, a car of mass 1 140 kg accelerates from rest. During the first few seconds of motion, the

krok68 [10]

Answer:

(a) $KE=16405.215 J$

(b) P = 6309.6981 W

(c) Value in above part is described as minimum because there would have been power loss in the actual system to achieve this acceleration from the state of rest.

Explanation:

Given:

mass of car, m = 1140 kg

expression of acceleration, $a=1.14t-0.210t^2+0.240t^3$

where "t" is time in seconds

initial time, $t_i=0 s$

final time, $t_f=2.6 s$

(a)

We know,

$\frac{dv}{dt} =a$

$dv=a.dt$

$v=\int\limits^{2.6}_0 {1.14t-0.210t^2+0.240t^3} \, dt$

$v=5.3648 m.s^{-1}$

Kinetic Energy

∴ $KE= \frac{1}{2} m.v^2$

$KE=\frac{1}{2}\times 1140\times 5.3648^2$

$KE=16405.215 J$

(b)

We know,

Power

$P= \frac{\Delta KE}{\Delta t}$

$P=\frac{16405.215}{2.6}$

P = 6309.6981 W

(c)

Value in above part is described as minimum because there would have been power loss in the actual system to achieve this acceleration from the state of rest.

3 0

2 years ago

Two objects attract each other gravitationally. If the mass of each object doubles, how does the gravitational force between the

madam [21]

Answer:

The gravitational force between them quadruples

Explanation:

According to law of gravitation, the force of attraction (F) between two masses m1 and m2 is directly proportional to the product of the masses and inversely proportional to the square of the distance(r) between them. Mathematically,

F1 = Gm1m2/r²... 1

If their masses doubles, the formula becomes;

F2 = G(2m1)(2m2)/r²

F2 = 4Gm1m2/r² ... 2

Dividing equation 2 by 1, we have;

F2/F1 = {4Gm1m2/r²}÷{Gm1m2/r²}

F2/F1 = 4Gm1m2/r²×r²/Gm1m2

F2/F1 = 4

F2 = 4F1

The gravitational force between the masses when they doubles quadruples.

6 0

2 years ago

A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g

Anna35 [415]

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

$Q_{cube}=Q_{water}$ (1)

But Q is related with mass (m), specific heat (c) and changes in temperature ( $\varDelta T$ )in the next way:

$Q=cm\varDelta T$ (2)

Using (2) on (1):

$c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}$

$(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

$(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

Because we have an equality and 0.385 < 4.186 then $\varDelta T_{cooper}>\varDelta T_{waterer}$ to conserve the equality

4 0

2 years ago

A solid, uniform disk of mass M and radius a may be rotated about an axis parallel to the disk axis, at variable distances from

cricket20 [7]

Answer:

the function varies linearly with the radius of the disk, so the smallest period is zero for a radius of zero centimeters

Explanation:

This system performs a simple harmonic movement where the angular velocity is given by

w = √ k / I

Where k is the constant recovered from the axis of rotation and I is the moment of inertia of the disk

The expression for the moment of inertia is

I = 1/2 m r²

Angular velocity, frequency and period are related

w = 2π f = 2π / T

Substituting

2π / T = √ k / I

T = 2π √ I / k

T = 2π √ (½ m r² / k)

T = (2π √m / 2k) r

We can see that the function varies linearly with the radius of the disk, so the smallest period is zero for a radius of zero centimeters

6 0

2 years ago