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tensa zangetsu [6.8K]
3 years ago
10

A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i

s the best approximation of the box’s acceleration? 120 m/s2 20 m/s2 1 m/s2 9 m/s2
Physics
1 answer:
Hoochie [10]3 years ago
8 0

The acceleration of the box is approximately 1 m/s^2

Explanation:

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

\sum F = ma

where

\sum F is the net force

m = 12.0 kg is the mass of the box

a is the acceleration

The net force can be written as

\sum F = F_a - F_f

where

F_a = 20 N is the applied forward force

F_f=9.0 N is the friction force

Combining the two equations,

F_a-F_f=ma

And solving for the acceleration,

a=\frac{F_a-F_f}{m}=\frac{20-9}{12}=0.9 m/s^2\sim 1 m/s^2

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel
trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

1\ mile=1609.34\ m

1\ h=3600\ seconds

18.03\times \frac{3600}{1609.34}=40.33\ mph

Speed of the boat by when it will hit the dock is 40.33 mph

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m

The distance at which the boat will have to start decelerating is 312.5 m

5 0
3 years ago
A guy wire 1005 feet long is attached to the top of a tower. When pulled taut, it touches level ground 552 feet from the base of
trasher [3.6K]

Answer:

56.7°

Explanation:

Imagine a rectangle triangle.

The triangle has 3 sides.

One side is the height of the tower, let's name it A.

Another side is the distance from the base of the tower to the point where the waire touches the ground. Let's name that B.

Sides A and B are perpendicular.

The other side is the length of the wire. Let's name it C.

From trigonometry we know that:

cos(a) = B / C

Where a is the angle between B anc C, between the wire and the ground.

Therefore

a = arccos(B/C)

a = arccos(552/1005) = 56.7°

7 0
3 years ago
X = 20 <br> v = 3.5 m/s <br> t = ?
N76 [4]

Answer:

16.5 I think

Explanation:

3 0
3 years ago
A student was walking with a mass of 40 kilograms and an acceleration of 1 m/s/s. What is the force exerted on her?
ValentinkaMS [17]

Answer:

B

Explanation:40x1=40       40=b

brainlyest plz

3 0
3 years ago
A calorimeter contained 350.0 g of water [cp=4.18 J/(g °C)] at 24.0 °C. An electric current was passed through a heater placed i
Shtirlitz [24]
Cp shows the amount of energy needed to raise temperature by one degree for one gram of water. 

Formula for calculating cp is:
cp= \frac{energy}{(mass)*( temperature_{change} ))}  \\ temperature_{change}= \frac{energy}{(mass)*( cp))}   \\  \\ temperature_{change}= \frac{16700}{(350)*( 4.18))}  \\  \\ temperature_{change}=2.73 \\  \\ temperature_{final} =temperature_{initial}+temperature_{change} \\ temperature_{final}=24 + 2.73 \\ temperature_{final}=26.73

Final temperature is 26.73°C.
4 0
3 years ago
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