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ss7ja [257]
3 years ago
13

What speed must a 600 kg car have in order to have the same momentum as a 1200 kg truck traveling at a velocity of 10 m/s to the

west? 5 m/s west B 20 mls west 5 m/s east 20 m/s east
A) 5m/s west
B) 20m/s west
C) 5m/s east
D) 20m/s east
​
Physics
1 answer:
yKpoI14uk [10]3 years ago
4 0

Answer:

20 m/s

Explanation:

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a 4kg box is resting on top of a shelf that is 2m high, what is the boxs gravitational potential relative to the floor​
Aloiza [94]

Answer:

80 J

Explanation:

PE = mgh

PE = (4 kg)(9.8 m/s^2)(2 m)

PE = 78.4 J and with sig figs, it would be 80 J

6 0
2 years ago
Read 2 more answers
A 12-volt automotive circuit has a current of 3 amps. Technician A says the electric power in this circuit is 36 watts. Technici
skelet666 [1.2K]

Answer:

Technician A is right.                                                    

Explanation:

Given that,

Voltage of circuit, V = 12 volt

Current in the circuit, I = 3 A

Technician A says the electric power in this circuit is 36 watts. Technician B says the electric power in this circuit is 4 watts. We need to say that which technician is correct.

The power of any circuit is given by :

P=V\times I

P=12\ V\times 3\ A

P = 36 watts

So, technician A is right. Hence, this is the required solution.

6 0
3 years ago
Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s
Ne4ueva [31]
The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
F=qvB \sin \theta
where \theta is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is q=2.7 \mu C=2.7 \cdot 10^{-6}C. In this case, v and B are perpendicular, so \theta=90^{\circ}, therefore we have:
B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T

2) In this second case, the angle between v and B is \theta=55^{\circ}. The charge is now q=42.0 \mu C=42.0 \cdot 10^{-6}C, and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N
5 0
3 years ago
Read 2 more answers
PLEASE HELP! I don't get it at all! Speed is one thing; distance is another. Where is the arrow you shoot up at 50m/s when it ru
LuckyWell [14K]
I got you b, V(final)^2=V(initial+2acceleration*displacement
So this turns to (0m/s)^2=(50m/s)^2+2(9.8)(d) so just flip it all around to isolate d so you get
-(50m/s)^2/2(9.8) = d so you get roughly 12.7555 meters up
4 0
3 years ago
Read 2 more answers
1*2+2+1+5 by how much?
kozerog [31]

Answer:

10

Explanation:

1*2+2+1+5=10

1x2+2+1+5=10

6 0
3 years ago
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