All categories

2 years ago

A boy is pulling a 150 kg crate with a force of 1150 N. If the crate experiences a frictional force of 490 N, find:

Physics

1 answer:

valentina_108 [34]2 years ago

7 0

<h3>For the aceleration:</h3>

First, let's find the resultant, and applicate 2nd law of Newton using the resultant, so:

R = ma

F - Ff = ma

Data:

F = Force = 1150 N

Ff = Friction force = 490 N

m = Mass = 150 kg

a = Aceleraction = ?

Replacing according our data:

1150 N - 490 N = 150 kg * a

660 N = 150 kg * a

660 N / 150 kg = a

a = 4,4 m/s² ← Aceleration of the object

<h3>For the normal force:</h3>

The normal force IS NOT the resultant force, the normal force's the force between the ground and the object, in another words, is the weight of the object, and for the weight:

w = mg

w = 150 kg * 10 m/s²

w = 1500 N ← Normal force between object and ground.

You might be interested in

A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50

laila [671]

Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

Given:

Height of the window=1.5 m
Time taken by ball to cover the window height=0.15

Now using equation of motion in one dimension we have

$s=ut+\dfrac{at^2}{2}$

Let u be the velocity of the ball when it reaches the top of the window

then

$1.5=0.15u+\dfrac{9.8\times0.15^2}{2}\\u=3.96\ \rm m/s\\$

Now u is the final velocity of the ball with respect to the top of the building

so let t be the time taken for it to reach the top of the window with this velocity

$3.96=gt\\t=0.4\ \rm s\\$

Let h be the height above the top of the window

$h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 0.4^2}{2}\\h=9.71\ \rm m$

3 0

2 years ago

Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 53 km/h. After six hour

Serhud [2]

For the answer to the question above, first find out the gradient.

m = rise/run 
=(y2-y1)/(x2-x1) 

the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" 
(x1,y1) = (3,53) 
(x2,y2) = (6,62) 

sub values back into the equation 
m = (62-53)/(6-3) 
m = 9/3 
m = 3 

now we use a point-slope form to find the the standard form 
y-y1 = m(x-x1) 
where x1 and y1 are any set of point given 
y-53 = 3(x-3) 
y-53 = 3x - 9 
y = 3x - 9 + 53 
y = 3x + 44 

y is the velocity of the car, x is the time.
I hope this helps.

4 0

3 years ago

A circular radar antenna on a Coast Guard ship has a diameter of 2.10 m and radiates at a frequency of 16.0 GHz. Two small boats

Anna35 [415]

Answer:

d = 76.5 m

Explanation:

To find the distance at which the boats will be detected as two objects, we need to use the following equation:

$\theta = \frac{1.22 \lambda}{D} = \frac{d}{L}$