Answer:
0 N
Explanation:
Applying,
F = qvBsin∅................. Equation 1
Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.
From the question,
Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°
Substitute these values into equation 1
F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)
Since sin0° = 0,
Therefore,
F = 0 N
Answer:
The current is 0.248 A
Explanation:
Given that,
Inductor 
Resistance 
Voltage = 15 volt
Time 
We need to calculate the current
Using formula of current

Where, V = voltage
R = resistance
L = inductance
T = time
Put the value into the formula


Hence, The current is 0.248 A.
Answer:
d=360 miles
Donna lives 360 miles from the mountains.
Explanation:
Conceptual analysis
We apply the formula to calculate uniform moving distance[
d=v*t Formula (1)
d: distance in miles
t: time in hours
v: speed in miles/hour
Development of problem
The distance Donna traveled to the mountains is equal to the distance back home, equal to d,then,we pose the kinematic equations for d, applying formula 1:
travel data to the mountains: t₁= 8 hours , v=v₁
d= v₁*t₁=8*v₁ Equation (1)
data back home : t₂=4hours , v=v₂=v₁+45
d=v₂*t₂=(v₁+45)*4=4v₁+180 Equation (2)
Equation (1)=Equation (2)
8*v₁=4v₁+180
8*v₁-4v₁=180
4v₁=180
v₁=180÷4=45 miles/hour
we replace v₁=45 miles/hour in equation (1)
d=8hour*45miles/hour
d=360 miles
Answer:
27.44 J
Explanation:
We can find the energy at the top of the slide by using the potential energy equation:
At the top of the slide, the swimmer has 0 kinetic energy and maximum potential energy.
The swimmer's mass is given as 7.00 kg.
The acceleration due to gravity is 9.8 m/s².
The (vertical) height of the water slide is 0.40 m.
Substitute these values into the potential energy equation:
- PE = (7.00)(9.8)(0.40)
- PE = 27.44
Since there is 0 kinetic energy at the top of the slide, the total energy present is the swimmer's potential energy.
Therefore, the answer is 27.44 J of energy when the swimmer is at the top of the slide.