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spin [16.1K]
4 years ago
12

Question #2

Physics
1 answer:
Contact [7]4 years ago
8 0

Answer:

Distance 20 km and Displacement 0 km

His displaceent is 0 km because he ends his walk where he started. The total distance of his walk is 20 km because he walks 10 km to the store + 10km back home.

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Both organisms attempt to use the same limited sources
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Active immunity is passed on from a mother to her baby.
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True.The immune system in babies. Antibodies are passed from mother to baby through the placenta during the last three months of pregnancy.
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In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight
algol13

Answer:

v_{f} =25m/s

Explanation:

Kinematics equation for constant acceleration:

v_{f}  =v_{o} + at=15+2*5=25m/s

4 0
3 years ago
The answer and how to do it?? Thanks
denis-greek [22]

Answer:

14 m/s²

Explanation:

Start with Newton's 2nd law: Fnet=ma, with F being force, m being mass, and a being acceleration. The applied forces on the left and right side of the block are equivalent, so they cancel out and are negligible. That way, you only have to worry about the y direction. Don't forget the force that gravity has the object. It appears to me that the object is falling, so there would be an additional force from going down from weight of the object. Weight is gravity (can be rounded to 10) x mass. Substitute 4N+weight in for Fnet and 1kg in for m.

(4N + 10 x 1kg)=(1kg)a

14/1=14, so the acceleration is 14 m/s²

4 0
3 years ago
Find the useful power output (in W) of an elevator motor that lifts a 2600 kg load a height of 30.0 m in 12.0 s, if it also incr
Annette [7]

Answer:

P = 251,916.667 W

Cost = 2,267.25 cents

Explanation:

To solve this question we will use the Work Energy Theorem, which is

W = dP + dK\\

Where

dP = Change in Potential Energy

dK = Change in Kinetic Energy

Change in Potential Energy

P_{i} = mgh_{i}\\  P_{f} = mgh_{f}

Where

P_{i} = Initial Potential Energy

P_{f} = Final Potential Energy

m = Mass of System = 10,000 kg

g = Acceleration due to gravity = 9.81 m/s

h_{i} = Initial Height = 0

h_{f} = Final Height = 30 m

Inputting the values we get the answer for dP

dP = P_{f} - P_{i}\\dP= mgh_{f} - mgh_{i}\\ dP= 10000(9.81)(30) - 0\\ dP= 2943000

Change in Kinetic Energy

K_{i} = \frac{1}{2} mv_{i} ^2\\ K_{f} = \frac{1}{2} mv_{f} ^2

Where

K_{i} = Initial Kinetic Energy

K_{f} = Final Kinetic Energy

m = Mass of System = 10,000 kg

g = Acceleration due to gravity = 9.81 m/s

v_{i} = Initial Velocity = 0 m/2

v_{f} = Final Velocity = 4 m/s

Inputting the values we get the answer for dK

dK = K_{f} - K_{i}\\ dK = \frac{1}{2} mv_{f} ^2 - \frac{1}{2} mv_{i} ^2\\ dK = \frac{1}{2} (10000)(4)^2 - 0 \\ dK = 80000

Total Work

W = dP + dK\\

Inputting the values

W = 2943000 + 80000

W = 3,023,000

a) Finding the useful Power Output

P = \frac{W}{t}

Where

P = Power Output

W = Work Done = 3,023,000J

t = Time = 12s

Inputting the values

P = \frac{3,023,000}{12}\\ P = 251,916.667

P = 251,916.667 W

b) Finding the Total Cost

Cost = $0.0900 x P/1000

Cost = $0.0900 x (251,916.667/1000)

Cost = $22.67 or 2,267.25 cents

4 0
3 years ago
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