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sweet-ann [11.9K]

3 years ago

12

An object moving at 36 km/h reduces its speed to 18 km/h in 2sec. If the mass of the object is 20 kg, find the average force act

ing on the object.

1 answer:

laiz [17]3 years ago

7 0

Answer:

-50 N

Explanation:

Givens:

V_i = 36 km/h

V_f = 18 km/h

t = 2 s

m = 20 kg

First we have to convert our km/h into m/s:

(36 km*(1000 m/1 km)) / (60 min *(60 s/1 min)) = 10 m/s

(18 km*(1000 m/1 km)) / (60 min *(60 s/1 min)) = 5 m/s

a = (V_f - V_i)/t

a = (5 m/s - 10 m/s) / 2 s

a = -2.5 m/s^2

F = m(a)

F = 20 kg(-2.5 m/s^2)

F = -50 N

It's a negative force meaning its acting on it opposite its current direction of movement.

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Higher mass protostars enter the main sequence: at the same rate, but at a higher luminosity and temperature. slower and at a lo

Morgarella [4.7K]

Answer:

<em>faster and at a higher luminosity and temperature.</em>

Explanation:

A protostar looks like a star but its core is not yet hot enough for fusion to take place. The luminosity comes exclusively from the heating of the protostar as it contracts. Protostars are usually surrounded by dust, which blocks the light that they emit, so they are difficult to observe in the visible spectrum.

A protostar becomes a main sequence star when its core temperature exceeds 10 million K. This is the temperature needed for hydrogen fusion to operate efficiently.

Stars above about 200 solar masses (Higher mass) generate power so furiously that gravity cannot contain their internal pressure. These stars blow themselves apart and do not exist for long if at all. A protostar with less than 0.08 solar masses never reaches the 10 million K temperature needed for efficient hydrogen fusion. These result in “failed stars” called brown dwarfs which radiate mainly in the infrared and look deep red in color. They are very dim and difficult to detect, but there might be many of them, and in fact they might outnumber other stars in the universe.

That is why higher mass protostars enter the main sequence at a <em>faster and at a higher luminosity and temperature.</em>

8 0

3 years ago

Compute the work performed when 32 pounds is lifted 10 feet.

Murljashka [212]

W = force * displacement
W = 32 pounds * 10 feet
Now you need to convert it to newton and meters
W = 142 N * 3.048 m = 434 J
(I approximated the conversions- I hope it helps)

7 0

3 years ago

A 44-turn rectangular coil with length ℓ = 17.0 cm and width w = 8.10 cm is in a region with its axis initially aligned to a hor

Mumz [18]

Answer:

The maximum induced emf in the rotating coil = 29.66V

The induced emf in the rotating coil when (t = 1.00 s) = 26.66V

The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s

Explanation:

Lets state the parameters we are being given right from the question:

Number of rectangular coil, (N) = 44

Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m

Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m

Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T

Angular Speed of Coil, (ω) = 64 rad/s

(a)

To calculate the induced emf in the rotating cell,we can use the formula:

emf = NBAωsin(ωt)

For maximum induced emf, the value of sin(ωt) will be 1

$emf_max$ = NBAω ; if (A = l × w) , we have:

$emf_max$ = NB(l × w)ω

subsitituting the parameters into the above equation; we have:

$emf_max$ = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64

= 29.66V

(b)

At t = 1s, the induced emf is calculated as:

emf = NBAωsin(ωt)

substituting the parameters into the equation, we have:

emf = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)

=26.66V

(c)

To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.

i.e $emf_max$ = $N\frac{d∅}{dt}$

since $emf_max$ = 29.66 and N = 44; we have:

29.66 = $44\frac{d∅}{dt}$

$\frac{d∅}{dt}$ = $\frac{29.66}{44}$

= 0.674 Wb/s

5 0

3 years ago

1.- La gráfica muestra la variación de la aceleración a de un objeto con el tiempo t.

STatiana [176]

Answer:

Te ayudo con una de prueba $

Explanation:

5 0

3 years ago

A heavy solid disk rotating freely and slowed only by friction applied at its outer edge takes 120 seconds to come to a stop.

alisha [4.7K]

Answer:

The time is 16 min.

Explanation:

Given that,

Time = 120 sec

We need to calculate the moment of inertia

Using formula of moment of inertia

$I=\dfrac{1}{2}MR^2$

If the disk had twice the radius and twice the mass

The new moment of inertia

$I'=\dfrac{1}{2}\times2M\times(2R)^2$

$I'=8I$

We know,

The torque is

$\tau=F\times R$

We need to calculate the initial rotation acceleration

Using formula of acceleration

$\alpha=\dfrac{\tau}{I}$

Put the value in to the formula

$\alpha=\dfrac{F\times R}{\dfrac{1}{2}MR^2}$

$\alpha=\dfrac{2F}{MR}$

We need to calculate the new rotation acceleration

Using formula of acceleration

$\alpha'=\dfrac{\tau}{I'}$

Put the value in to the formula

$\alpha=\dfrac{F\times R}{8\times\dfrac{1}{2}MR^2}$

$\alpha=\dfrac{2F}{8MR}$

$\alpha=\dfrac{\alpha}{8}$

Rotation speed is same.

We need to calculate the time

Using formula angular velocity

$\Omega=\omega'$

$\alpha\time t=\alpha'\times t'$

Put the value into the formula

$\alpha\times120=\dfrac{\alpha}{8}\times t'$

$t'=960\ sec$

$t'=16\ min$

Hence, The time is 16 min.

5 0

3 years ago