Answer: current I = 0.96 Ampere
Explanation:
Given that the
Resistance R = 60 Ω
Power = 55 W
Power is the product of current and voltage. That is
P = IV ...... (1)
But voltage V = IR. From ohms law.
Substitutes V in equation (1) power is now
P = I^2R
Substitute the above parameters into the formula to get current I
55 = 60 × I^2
Make I^2 the subject of formula
I^2 = 55/60
I^2 = 0.92
I = sqr(0.92)
I = 0.957 A
Therefore, 0.96 A current must be applied.
Answer:
A. Remove everything in the refrigerator to lighten the load.
B. Put a lubricant between the surface of the object and the floor
C. Use round objects, like pencils , to decrease the friction and push the refrigerator over the pencils more easily
Explanation:
Force of friction is a resistance force which acts between two surfaces which are in relative motion. Friction is both boon and bane. Due to friction, we are able to sit, walk etc but also, due to friction there is dissipation of energy. Friction can be reduced by applying lubricants, reducing contact area, reducing the load.
F = μN where N is the normal force which depends on the mass.
Thus, by reducing the load, force of friction can be reduced. Round objects like wheels can also be used. By this the contact area reduces.
Answer:
Matter is anything that has mass and takes up space. A mixture is matter that can vary in composition.....because, the substances in a heterogeneous mixture are not evenly mixed, two samples of the same mixture can have different amounts of the substance....
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Answer:
the mass of water is 0.3 Kg
Explanation:
since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:
Q water + Q copper = Q surroundings =0 (insulated)
Q water = - Q copper
since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature
and denoting w as water and co as copper :
m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) = m co * c co * (T co - Ti eq)
m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]
We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C
if we assume that both specific heats do not change during the process (or the change is insignificant)
m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]
m w= 1.80 kg * 0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))
m w= 0.3 kg
Q: ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself off in the opposite direction at 60 m/s. What is the magnitude of the average force the wall exerts on Ken during this rapid maneuver
Answer:
-6461.54 N
Explanation:
From Newton's Fundamental equation,
F = m(v-u)/t.................... Equation 1
Where F = Force exerted in sonic, m = mass of ken, v = final velocity, u = initial velocity, t = time.
Given: m = 0.75 kg, v = - 60 m/s (opposite direction), u = 52 m/s, t = 13 ms = 0.013 s
Substitute into equation 1
F = 0.75(-60-52)/0.013
F = 0.75(-112)/0.013
F = -84/0.013
F = -6461.54 N
Note: The negative sign tells that the force act in opposite direction to the initial motion of ken.
Hence the magnitude of the average force of the wall = -6461.54 N
