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lina2011 [118]
3 years ago
15

Which equation is true of an atom with no electrical charge?

Physics
2 answers:
Svetach [21]3 years ago
7 0

Answer:

c

Explanation:

just did the test on A P E X

lord [1]3 years ago
6 0

Answer: c

Explanation:

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If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is
Leokris [45]
73 Newton is the correct answer
6 0
3 years ago
Is playing baseball work
Arisa [49]

Answer:

Depends

Explanation:

If you play baseball on a team then yes if you don't and you play on your free time then no it isn't work because your doing it for fun now if you do it on a team then you are working with your team and winning as a team

8 0
3 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
2. An object with a mass of 3.8 kg has four forces acting on it.
marshall27 [118]

Answer:4.7N

Explanation:both of the sides are equal

7 0
3 years ago
Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit
Ede4ka [16]

Answer:

386.2^{\circ}F

Explanation:

We are given that

P_1=200lbf/in^2

P_2=60lbf/in^2

v_1=200ft/s

v_2=1700ft/s

T_1=500^{\circ}F

Q=0

C_p=1BTU/lb^{\circ}F

We have to find the exit temperature.

By steady energy flow equation

h_1+v^2_1+Q=h_2+v^2_2

C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}

1BTU/lb=25037ft^2/s^2

Substitute the values

1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}

500+1.598=T_2+115.4

T_2=500+1.598-115.4

T_2=386.2^{\circ}F

7 0
4 years ago
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