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2 years ago

A car of mass 1100kg moves at 24 m/s. What is the braking force needed to bring the car to a halt in 2.0 seconds? N

Physics

1 answer:

LenaWriter [7]2 years ago

8 0

13200N

Explanation:

Given parameters:

Mass = 1100kg

Velocity = 24m/s

time = 2s

unknown:

Braking force = ?

Solution:

The braking force is the force needed to stop the car from moving.

Force = ma = $\frac{mv}{t}$

m is the mass of the car

v is the velocity

t is the time taken

Force = $\frac{1100 x 24}{2}$ = 13200N

Learn more:

Force brainly.com/question/4033012

#learnwithBrainly

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babunello [35]

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

$y=\frac{m\lambda D}{d}$

At m = 0

$y_0=0$

$y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm$

At m = 1

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The slit separation is 0.00001266 m

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2 years ago

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STALIN [3.7K]

Answer:

-5.0m/s

Explanation:

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2 years ago

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Please help! will give brainlest!!!!!!!!!!!!

eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

$a=0$

Therefore the net force is zero as well:

$\sum F = 0$

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

$\sum F = F cos \theta - F_f = 0$

where

$F cos \theta$ is the horizontal component of the applied force, with

F = 22.5 N

$\theta=32.0^{\circ}$

$F_f$ is the force of friction

And solving for $F_f$ , we find

$F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N$

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2 years ago

Grain is pored into a silo to be stored for later use. Due to the friction between pieces of grain as they rub against eachother

MariettaO [177]

Answer:

3.1×10⁻¹¹ N

Explanation:

Use Coulomb's law:

F = k q₁ q₂ / r²

F = (9×10⁹) (6.0×10⁻¹⁰) (2.3×10⁻¹⁵) / (0.02 m)²

F = 3.1×10⁻¹¹

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2 years ago

A falling skydiver opens his parachute. A short time later, the weight of the skydiver-parachute system and the drag force exert

Dimas [21]

A falling skydiver opens his parachute. A short time later, the weight of the skydiver-parachute system and the drag force exerted on the system are equal in magnitude. The following statements predicts the motion of the skydiver at this time

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Explanation:

Immediately on leaving the aircraft, the skydiver accelerates downwards due to the force of gravity. There is no air resistance acting in the upwards direction, and there is a resultant force acting downwards. The skydiver accelerates towards the ground.

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The net force and the acceleration on the falling skydiver is upward.

An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down.

As the speed decreases, the amount of air resistance also decreases until once more the skydiver reaches a terminal velocity.

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2 years ago