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boyakko [2]
3 years ago
11

A 20.0 kg mass moving at a velocity of + 3.0 m/s is stopped by a constant force of 15.0 n. how many seconds must the force act o

n the mass to stop it? html editor keyboard shortcuts
Physics
1 answer:
slamgirl [31]3 years ago
4 0
The mass is 20 kg, initial speed is 30 m/s and the final speed is 0 m/s (rest), 
while the force is 15 N
Using second newton's law of motion which states that the rate of change in momentum is directly proportional to the resultant force. Then
 Ft = MV-MU
15 t = 20 (0-3)
15t = -60
  t = -4 sec ( negative because the force is acting on opposite direction to the motion)
Therefore, the time taken is 4 seconds
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How does the balanced chemical equation show the conservation of matter in the chemical reaction?
Dmitry [639]

C. The number of F atoms in the reactants equals the number of F atoms in the products.

4 0
2 years ago
An automobile with an initial speed of 4.92 m/s accelerates uniformly at the rate of 3.2 m/s2 . Find the final speed of the car
Rudik [331]

Answer:19.32 m/s

Explanation:

Given

initial speed of car(u)=4.92 m/s

acceleration(a)=3.2 m/s^2

Speed of car after 4.5 s

using equation of motion

v=u+at

v=4.92+3.2\times 4.5=4.92+14.4

v=19.32 m/s

Displacement of the car after 4.5 s

v^2-u^2=2as

19.32^2-4.92^2=2\times 3.2\times s

349.05=2\times 3.2\times s

s=54.54 m

4 0
3 years ago
After observing the electric field in your trials above, where was the electric field the strongest? what was the direction of t
earnstyle [38]
<span>Answer: Answer is The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source charge.</span>
8 0
3 years ago
A shopper pushes a 5.32 kg grocery cart with a 12.7 N force directed at -28.7° below horizontal. A friction force of 8.33 N push
atroni [7]

Answer:

0.8214 m/s^2

Explanation:

Fnet= Fpushed - Ffriction

Fpushed = 12.7N      Ffriction = 8.33N

Fnet = 12.7N - 8.33N = 4.37N

Fnet= mass(acceleration)

Fnet = 4.37N    mass = 5.32 kg

4.37N = 5.32 kg(acceleration)

acceleration= 0.8214 m/s^2

3 0
3 years ago
A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t
Jlenok [28]

Answer:

14.36 N

Explanation:

T_{1} = Tension in string 1

T_{2} = Tension in string 2

m_b = mass of the bar = 2.7 kg

W_b = weight of the bar

weight of the bar is given as

W_b = m_{b} g = (2.7) (9.8) = 26.46N

m_m = mass of the bar = 1.35 kg

W_m = weight of the monkey

weight of the monkey is given as

W_m = m_{m} g = (1.35) (9.8) = 13.23N

Using equilibrium of torque about left end

W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33 N

Using equilibrium of force in vertical direction

T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36 N

7 0
3 years ago
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