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Bad White [126]
3 years ago
11

What is the definition of a point object?

Physics
1 answer:
Kaylis [27]3 years ago
4 0
Point object is an expression that is used in Kinematics. It refers the object whose dimensions are ignored or neglected and is treated as a dot object to simplify the calculations. When the object is treated as a point object it means the whole mass is summed in one point, you can imagine it as a sphere which has the mass and the radius is 0...
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While a roofer is working on a roof that slants at 39.0 degrees above the horizontal, he accidentally nudges his 88.0 N toolbox,
Ostrovityanka [42]

Answer:

V= 6.974 m/s

Explanation:

Component( box) weight acting parallel and down roof 88(sin39.0°)=55.4 N

Force of kinetic friction acting parallel and up roof = 18.0 N

Fnet force acting on tool box acting parallel and down roof

Fnet= 55.4 - 18.0

Fnet=37.4 N

acceleration of tool box down roof

a = 37.4(9.81)/88.0

a= 4.169 m/s²

d = 4.90 m

t = √2d/a

t= √2(4.90)/4.169

t= 1.662 s

V = at

V= 4.169(1.662)

V= 6.974 m/s

5 0
3 years ago
Help plzzzzzzzzzzzzzzzz!
aksik [14]
 

concave <span>ray diagrams were constructed in order to determine the general location, size, orientation, and type of image formed by concave mirrors. Perhaps you noticed that there is a definite relationship between the image characteristics and the location where an object placed in front of a concave mirror. but, convex</span><span>ray diagrams were constructed in order to determine the location, size, orientation, and type of image formed by concave mirrors. The ray diagram constructed earlier for a convex mirror revealed that the image of the object was virtual, upright, reduced in size and located behind the mirror. </span>
3 0
3 years ago
A substance freezes at -58c therefore substance melts at ?
Yanka [14]
The freezing point is the same as the melting point.

If it freezes at -58°C, hence the melting point is also <span>-58°C.</span>
3 0
3 years ago
A heat pump with a COP of 3.15 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the te
Jet001 [13]

Answer: 1026s, 17.1m

Explanation:

Given

COP of heat pump = 3.15

Mass of air, m = 1500kg

Initial temperature, T1 = 7°C

Final temperature, T2 = 22°C

Power of the heat pump, W = 5kW

The amount of heat needed to increase temperature in the house,

Q = mcΔT

Q = 1500 * 0.718 * (22 - 7)

Q = 1077 * 15

Q = 16155

Rate at which heat is supplied to the house is

Q' = COP * W

Q' = 3.15 * 5

Q' = 15.75

Time required to raise the temperature is

Δt = Q/Q'

Δt = 16155 / 15.75

Δt = 1025.7 s

Δt ~ 1026 s

Δt ~ 17.1 min

5 0
3 years ago
Read 2 more answers
A ray diagram shows that an object is placed in front of a plane mirror. What are the characteristics of the image produced by t
romanna [79]
Same size as object should be the answer, it is a “plane” mirror
3 0
2 years ago
Read 2 more answers
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