Given Information:
Inclined plane length = 8 m
Inclined plane height = 2 m
Weight of ice block = 300 N
Required Information:
Force required to push ice block = F = ?
Answer:
Force required to push ice block = 75 N
Explanation:
The force required to push this block of ice on a inclined plane is given by
F = Wsinθ
Where W is the weight of the ice block and θ is the angle as shown in the attached image.
Recall from trigonometry ratios,
sinθ = opposite/hypotenuse
Where opposite is height of the inclined plane and hypotenuse in the length of the inclined plane.
sinθ = 2/8
θ = sin⁻¹(2/8)
θ = 14.48°
F = 300*sin(14.48)
F = 75 N
Therefore, a force of 75 N is required to push this ice block on the given inclined plane.
Answer:
Explanation:
Given that
re= 46 cm
Vp= 180 m/s
We know that
So
Now by putting the all given values in the questions
So the average electric field is .
<span>14 m/s
Assuming that all of the energy stored in the spring is transferred to dart, we have 2 equations to take into consideration.
1. How much energy is stored in the spring?
2. How fast will the dart travel with that amount of energy.
As for the energy stored, that's a simple matter of multiplication. So:
20 N * 0.05 m = 1 Nm = 1 J
For the second part, the energy of a moving object is expressed as
KE = 0.5 mv^2
where
KE = Kinetic energy
m = mass
v = velocity
Since we now know the energy (in Joules) and mass of the dart, we can substitute the known values and solve for v. So
KE = 0.5 mv^2
1 J = 0.5 0.010 kg * v^2
1 kg*m^2/s^2 = 0.005 kg * v^2
200 m^2/s^2 = v^2
14.14213562 m/s = v
So the dart will have a velocity of 14 m/s after rounding to 2 significant figures.</span>
Answer:
Show attached picture
Explanation:
Let's call V the voltage provided by the battery in the circuit. M is the multimeter (let's call 
its internal resistance) and R indicates the resistance of the light bulb.
We know that the meter's internal resistance is 1000 times higher than the bulb's resistance:
(1)
Both the meter and the bulb are connected in parallel to the battery, so they both have same potential difference at their terminals:
Using Ohm's law, 
, we can rewrite the previous equation as:
where
is the current in the meter
is the current in the bulb
Using (1), this equation becomes
so, the current in the meter is 1000 times less than through the bulb.
