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VMariaS [17]
3 years ago
14

A proton travels with a speed of 4.2×106 m/s at an angle of 30◦ west of north. A magnetic field of 2.5 T points to the north. Fi

nd the magnitude of the magnetic force on the proton. (The magnetic force experienced by the proton in the magnetic field is proportional to the component of the proton’s velocity that is perpendicular to the magnetic field.)
Physics
1 answer:
Arada [10]3 years ago
6 0

Answer:

8.4\cdot 10^{-13} N

Explanation:

The magnitude of the magnetic force on the proton is given by:

F=qvB sin \theta

where:

q=1.6\cdot 10^{-19} C is the proton charge

v=4.2\cdot 10^6 m/s is the proton velocity

B=2.5 T is the magnetic field

\theta=30^{\circ} is the angle between the direction of v and B

Substituting into the formula, we find

F=(1.6\cdot 10^{-19}C)(4.2\cdot 10^6 m/s)(2.5 T) sin 30^{\circ}=8.4\cdot 10^{-13} N

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A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci
vovikov84 [41]
According to the law of conservation of momentum:

m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know. 

m1 = 125 kg     v1 = 12m/s      v'1 = -12.5m/s
m2 = 235kg      v2 = -13m/s     v'2 = ?

m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'
(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'
1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')
-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation. 

-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')
7.5kg.m/s=(235kg)(v_{2}')
(7.5kg.m/s)/(235kg)=(v_{2}')
0.03m/s=(v_{2}')

The velocity of the 2nd car after the collision is 0.03m/s.
5 0
3 years ago
An electric Kettle is rated at 25 W. Calculate the quantity of heat generated in 2s
MatroZZZ [7]

Answer:

Energy consumed by the electric kettle in 9.5 min =Pt=(2.5×10

3

)×(9.5×60)=14.25×10

5

J

Energy usefully consumed =msΔT=3×(4.2×10

3

)×(100−15)=10.71×10

5

where s=4.2J/g

o

C= specific heat of water and boiling point temp=100

o

C

Heat lost =14.25×10

5

−10.71×10

5

=3.54×10

5

4 0
3 years ago
Which of the following intermolecular forces explains why iodine (I2) is a solid at room temperature?
egoroff_w [7]
"Dispersion forces" is the one intermolecular force among the following choices given in the question that <span>explains why iodine (I2) is a solid at room temperature. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that the answer has helped you.</span>
3 0
3 years ago
Stairway must have uniform riser height and tread depth; variations in riser height or tread depth shall not be over _______ inc
alexandr1967 [171]

Answer:

\frac{3}{8} inches

Explanation:

the variations in riser height or tread depth should not be grater than \frac{3}{8} inches that is equal to 9.5 mm but the maximum riser height should be the  \frac{81}{4} inch  but variation in riser height should not exceed to \frac{3}{8} inches. The minimum riser height should be 7 inches which is equal to the 178 mm

5 0
3 years ago
A commuter train travels from New York to Washington, DC, and back in 6 hours and 5 minutes. The distance between the two statio
vovikov84 [41]

Answer:

The train's displacement is zero.

Explanation:

Given data,

The time taken by the train from NY to Washington and back is, t = 6 h 5 min

The distance between the two stations is, d = 363 km

Therefore, the total distance the train traveled is, d' = 726 km

The displacement is defined as the change in position coordinates with respect to its original position.

If the train travels from one point and returns back to the same point after some time, there is no change in the position coordinates with respect to its original position.

Hence, the train's displacement is zero.

3 0
3 years ago
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