Answer:
If using radians: 16.2
If using degrees: 12.8
Don't forget your units
Explanation:
I assume that x represents the velocity?
If so, then just substitute the value t and solve
TheThe distance they have covered for trip is 219Km.
<h3>What do you mean by uniformly accelereted motion?</h3>
When an object is traveling in a straight line with an increase in velocity at equal intervals of time.
The total time for the trip is
T→ t1+ 22 min = t1+ 0.367 h ,where t1 is the time spent traveling at
V1= 89.5 km/ h .
the distance traveled is ∆x = V1t1=Vavg T
after applying value and calculating it we get
t1= 2.44h for a total time of
t total we get 2.81 h .
∆x =V1T1 = VavgTtotal
∆x = 77×2.81= 219 Km
to learn more about Uniform accelereted motion click here brainly.com/question/12920060
#SPJ4
The magnetic field is .
<h3>What is Magnetic Field?</h3>
The area in which the force of magnetism works around a magnetic substance or a moving electric charge is known as the magnetic field. a diagram that shows the magnetic field and how a magnetic force is distributed within and outside of a magnetic substance.
a) the magnetic field, B outside the wire
Using ampere law:
so:
the direction is in
b) the magnetic field, B inside the wire
Using ampere law:
so:
the direction is in
to learn more about magnetic field go to - brainly.com/question/7802337
#SPJ4
Answer:
(I). The store energy is 0.432 J.
(II). The energy density is 687.54 J/m³.
Explanation:
Given that,
Voltage = 12.0 V
Capacitor = 6.00 mf
Radius = 0.1 m
Distance = 10 mm
(I). We need to calculate the energy store
Using formula of store energy
Where, C = capacitor
V = voltage
Put the value into the formula
(II). We need to calculate the energy density
Using formula of energy density
Where, r = radius
d = distance
Put the value into the formula
Hence, (I). The store energy is 0.432 J.
(II). The energy density is 687.54 J/m³.
Answer:
671.76 kg or 6590 N
Explanation:
So the buoyant force generated by the floating ice is equals to the mass of water displaced by the submerged ice. We also need to account for gravity of ice. The resulting additional mass that the ice sheet can support is the difference between the mass of water displaced by ice and the mass of ice submerged totally in water.
So the ice piece can support an additional 671.76 kg of bear, or 671.76 * 9.81 = 6590 N