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SSSSS [86.1K]
4 years ago
15

How has two dogs go awnser my other qestion first one will get brainly anser

Physics
1 answer:
aksik [14]4 years ago
7 0
Yes, yea, yep, you don't make much sense.

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What is f if you have an object 2.0 m from the concave mirror, and the image is 4.0 m from the mirror?
dezoksy [38]

Answer:

the f is 1.3 m

Explanation:

Given that

An object 2.0m is from the concave mirror

And, the image is 4.0m from the mirror

We need to find out the f

So,

u = -2 m

v = -4 m

1 ÷ f = 1 ÷ v + 1 ÷ u

= (-1 ÷ 4) + (-1 ÷ 2)

= -3 ÷4

f = -4 ÷ 3

= |1.3| m

Hence, the f is 1.3 m

7 0
3 years ago
Can someone help me please? I’m so confused
Svetradugi [14.3K]

Answer:

what do you need help with? Like which type of print each image is?

Explanation:

5 0
3 years ago
What is the direction of the electric force exerted by paperclip 1 on paperclip 2?
Tema [17]
In general, the electric force in an electric field is exerted outward from a positive atom, and inward for the negative atom. Therefore based on the figure you specified, we can say that the <span>electric force exerted by paperclip 1 on paperclip 2 is repulsive.</span>
6 0
4 years ago
A railroad car of mass 3.45 ✕ 104 kg moving at 2.60 m/s collides and couples with two coupled railroad cars, each of the same ma
Alex

Answer:

a) 1.67 m/s

b) 23kJ

Explanation:

We need to apply the linear momentum conservation formula, that states:

m1*v_{o1}+m2*v_{o2}=m1*v_{f1}+m2*v_{f2}

in this case:

3.45*10^4kg*2.60m/s+2*3.45*10^4kg*1.20m/s=3*m1*v_{f}\\v_f=1.67m/s

the initital kinetic energy is:

K_i=\frac{1}{2}*3.45*10^4kg*(2.60m/s)^2+2(\frac{1}{2}*3.45*10^4kg*(1.20m/s)^2\\K_i=167kJ

and the final:

K_f=3*\frac{1}{2}*3.45*10^4kg*(1.67m/s)^2\\K_f=144kJ

The energy lost is given by:

E_l=|K_f-K_i|\\E_l=23kJ

8 0
3 years ago
A ball is thrown vertically upwards at 19.6 m/s. For its complete trip (up and back down to the starting position), its average
stira [4]

When the ball reaches its original position, it will have the same speed (but would be traveling in the opposite direction). So the average speed is

\dfrac{19.6\,\frac{\mathrm m}{\mathrm s}-19.6\,\frac{\mathrm m}{\mathrm s}}{\Delta t}=0

regardless of how long the ball was in the air.

5 0
3 years ago
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