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beks73 [17]
2 years ago
7

A 1.23 x 10^3 kilogram car is traveling east at 25 meters per second. The brakes are applied and the car is brought to rest in 5

.00 seconds. Calculate the magnitude of the total impulse applied to the car to bring it to rest. State the direction of impulse applied to the car.
Physics
1 answer:
ahrayia [7]2 years ago
3 0

Answer:

\text{Magnitude: }30,000\:\text{Ns},\\\text{Direction: opposite direction of car's movement}}

Explanation:

*Edit: The original question states a mass of 1.23\cdot 10^3\:\text{kg}. Since the, the poster has corrected it to 1.20\cdot 10^3\:\text{kg} and therefore the answers have been change to account for the typo.

The impulse-momentum theorem states that the impulse on a object is equal to the change in momentum of the object.

Therefore, we have the following equation:

F\Delta t=\Delta p, where F\Delta t is impulse (another way to find impulse) and \Delta p is change in momentum.

Because the car is being slowed to a rest, its final velocity will be zero, and therefore its final momentum will also be zero. Since momentum is given as p=mv, the car's change in momentum is 1.20\cdot 10^3\cdot 25-0=1230\cdot 25=30,000\:\text{kgm/s}.

As we wrote earlier, this is also equal to the magnitude of impulse on the object. The time it takes to stop the car is actually irrelevant to finding the total impulse. However, if we were to calculated the average applied force on the car, we would need how long it takes to bring it to rest (refer to F\Delta t).

The direction of the impulse must be exactly opposite to the car's direction, since we are slowing it to a stop.

Therefore, the impulse on the car is \boxed{30,750\:\text{Ns, opposite direction of car's movement}}.

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A ball is thrown straight up from a bridge at a speed of 11.0 m/s. If it takes 5.5 seconds to hit the water below, what is the v
seropon [69]

Answer:

 v = 42.92 m/s

Explanation:

Given,

initial speed of the ball, v = 11 m/s

time taken to hit the ground = 5.5 m/s

velocity of the ball just before it hit the ground, v = ?

time taken by the ball to reach the maximum height

using equation of motion

 v = u + at

final velocity = 0 m/s  

 0 = 11 - 9.8 t

  t = 1.12 s.

time taken by the ball to reach the water from the maximum height

 t' - 5.5 -1.12 = 4.38 s

using equation of motion for the calculation of speed just before it hit the water.

 v  = u + a t

 v = 0 + 9.8 x 4.38

 v = 42.92 m/s

Velocity of the ball just before it reaches the water is equal to v = 42.92 m/s

3 0
3 years ago
Calculate how far a ball will travel horizontally if the ball reaches a high of 1.5 m above the ground and is thrown at a horizo
nataly862011 [7]

67.5

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6 0
3 years ago
A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
mamaluj [8]

The vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

<h3>Tension in the cable</h3>

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (29 x 9.8)/(2 x sin57)

T = 169.43 N

<h3>Vertical component of the force</h3>

T + F = W

F = W - T

F = (9.8 x 29) - 169.43

F = 114.77 N

Thus, the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

Learn more about tension here: brainly.com/question/24994188

#SPJ1

6 0
2 years ago
A battery has a potential resistance of 12 V and a current of 1280 mA. What is the resistance? PLS HELP ME ASAPPPPPPPPPP
Romashka-Z-Leto [24]

Resistance (R) = <u>9.375 ohm (Ω)</u>

Steps:

R =   \frac{V}{I}

=    \frac{12 \: volt}{1280 \:  milliampere}

=  \frac{12 \: volt}{1.28 \: ampare}

= 9.375 \: ohm (Ω)

6 0
3 years ago
Energy Conservation With Conservative Forces: If a spring-operated gun can shoot a pellet to a maximum height of 100 m on Earth,
crimeas [40]

Answer:

h' = 603.08 m

Explanation:

First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)

h = height of pellet = 100 m

Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)

Vi = Initial Velocity of Pellet = ?

Therefore,

(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²

Vi = √(1960 m²/s²)

Vi = 44.27 m/s

Now, we use this equation at the surface of moon with same initial velocity:

2g'h' = Vf² - Vi²

where,

g' = acceleration due to gravity on the surface of moon = 1.625 m/s²

h' = maximum height gained by pellet on moon = ?

Therefore,

2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²

h' = (1960 m²/s²)/(3.25 m/s²)

<u>h' = 603.08 m</u>

4 0
3 years ago
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