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1 year ago

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A 2300-kg car slows down at a rate of 3.0 m/s2 when approaching a stop sign. What is the magnitude of the net force causing it t

o slow down?

1 answer:

weeeeeb [17]1 year ago

4 0

The magnitude of the net force causing the 2300kg car to slow down is 6900N

HOW TO CALCULATE FORCE:

The net force applied on a moving object can be calculated by multiplying the mass of the object by its acceleration. That is;

Force (N) = mass (kg) × acceleration (m/s²)

According to this question, a 2300-kg car slows down at a rate of 3.0 m/s2 when approaching a stop sign. The net force causing the car to stop can be calculated as follows:

F = 2300kg × 3m/s²

F = 6900N

Therefore, the magnitude of the net force causing the 2300kg car to slow down is 6900N.

Learn more at: brainly.com/question/18109210?referrer=searchResults

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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 30 m/s2 for 35 s , then ru

Gre4nikov [31]

Answer:

a) $h_m=74625\ m$

b) $t=265.55\ s$

Explanation:

Given:

mass of rocket, $m_r=200\ kg$

mass of fuel, $m_f=100\ kg$

acceleration of the rocket consuming fuel, $a=30\ m.s^{-2}$

time after which the fuel exhaust, $t_f=35\ s$

<u>During the phase of fuel exhaustion:</u>

<u>velocity attained by the rocket just as the fuel ends:</u>

$v_f=u+a.t_f$

where:

$u=$ initial velocity of the rocket = 0

$v_f=0+30\times 35$

$v_f=1050\ m.s^{-1}$ this will be the initial velocity for the phase of ascending of the rocket's height under the influence of gravity.

<u>height at which the fuel finishes:</u>

$v_f^2=u^2+2a.h_f$

$1050^2=0^2+2\times 30\times h_f$

$h_f=18375\ m$

<u>During the phase of ascend in height of rocket after the fuel is over:</u>

<u>Time taken to reach the top height after the fuel is over:</u>

$v=v_f+g.t'$

at top v = (final velocity during this course of motion )= 0 $m.s^{-1}$

$0=1050-9.8\times t'$

$t'=107.1429\ s$

<u>Height ascended by the rocket after the fuel is over:</u>

$v^2=v_f^2+2g.h'$

at the top height the velocity is zero

$0^2=1050^2-2\times 9.8\times h'$ (-ve sign denotes that the direction of motion is opposite to that of acceleration)

$h'=56250\ m$

<u>Therefore the maximum altitude attained by the rocket:</u>

$h_m=h_f+h'$

$h_m=18375+56250$

$h_m=74625\ m$

b)

time taken by the rocket to fall back to the earth:

$h_m=v.t_m+\frac{1}{2} g.t_m^2$

where:

$v=$ initial velocity of the rocket during the course of free fall from the top height.

$74625=0+4.9\times t_m^2$

$t_m=123.41\ s$

Now the total time for which the rocket is in the air:

$t=t_f+t'+t_m$

$t=35+107.1429+123.41$

$t=265.55\ s$

4 0

2 years ago

A baseball is dropped off the side of a cliff. It free-falls to the ground in 8 seconds. During the third second, the ball is tr

kozerog [31]

The acceleration of gravity is 9.8 m/s² . That means that a falling object
is always falling 9.8 m/s faster than it was falling 1 second earlier.

If an object is not slowed by air resistance, and has far enough to go
so that it's still falling after three whole seconds, then at the end of
three seconds it's falling at

(9.8 m/s²) x (3 sec) = 29.4 m/s

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2 years ago

Read 2 more answers

A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The

mestny [16]

Answer:

a. Δx = 2.59 cm

Explanation:

mb = 0.454 kg , mp = 5.9 x 10 ⁻² kg , vp = 8.97 m / s , k = 21.0 N / m

Using momentum conserved

mb * (0) + mp * vp = ( mb + mp ) * vf

vf = ( mp / mp + mb) * vp

¹/₂ * ( mp + mb) * (mp / mp +mb) ² * vp ² = ¹/₂ * k * Δx²

Solve to Δx '

Δx = √ ( mp² * vp² ) / ( k * ( mp + mb )

Δx = √ ( ( 5.9 x 10⁻² kg ) ² * (8.97 m /s) ² / [ 21.0 N / m * ( 5.9 x10 ⁻² kg + 0.454 kg ) ]

Δx = 0.02599 m ⇒ 2.59 cm

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2 years ago

During which phase of mitosis do the chromosomes line up across the center of the cell?

nirvana33 [79]

Metaphase; the centromeres of duplicated chromosomes line up in the middle of the cell. (It's also the shortest phase of mitosis).

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2 years ago

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1050 kg and was approaching at

Soloha48 [4]

Answer:

v = 11.0 m/s at 198.6° (18.6° south of west)

ΔKE = -145 kJ

Explanation:

I assume you want to find the final velocity and the change in kinetic energy.

Take east to be +x and north to be +y.

Momentum is conserved in the x direction:

(1050 kg) (0 m/s) + (750 kg) (-25.0 m/s) = (1050 kg + 750 kg) vₓ

vₓ = -10.4 m/s

Momentum is conserved in the y direction:

(1050 kg) (-6.00 m/s) + (750 kg) (0 m/s) = (1050 kg + 750 kg) vᵧ

vᵧ = -3.50 m/s

The magnitude of the final velocity is:

v² = (-10.4 m/s)² + (-3.50 m/s)²

v = 11.0 m/s

The direction of the final velocity is:

θ = atan(-3.50 m/s / -10.4 m/s)

θ = 198.6°

The initial kinetic energy is:

KE₀ = ½ (1050 kg) (6.00 m/s)² + ½ (750 kg) (25.0 m/s)²

KE₀ = 253,275 J

The final kinetic energy is:

KE = ½ (1800 kg) (11.0 m/s)²

KE = 108,682 J

The change in kinetic energy is:

ΔKE = 108,682 J − 253,275 J

ΔKE ≈ -145,000 J

7 0

2 years ago