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fiasKO [112]
3 years ago
15

Can anyone please help me on this Science/Physics question ( Number 5) (DUE TOMORROW)

Physics
1 answer:
Alecsey [184]3 years ago
3 0
I think it’s this: a) litres
b) millilitres
c)kg
d) mg
e)g
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At the same moment, one rock is thrown upward at 4.5 m/s and another thrown downward at 6.2 m/s. What is the relative velocity o
erastova [34]
The correct answer is 
<span>C) -10.7 m/s 

In fact, the first rock is moving upward with velocity +4.5 m/s, while the second rock is moving downward with velocity -6.2 m/s, with respect to a fixed reference frame. In the reference frame of the first rock, instead, the second rock is moving with velocity equal to its velocity in the fixed frame minus the velocity of the reference frame of the first rock:
</span>v=-6.2 m/s -(+4.5 m/s) = -10.7 m/s<span>
</span>
8 0
4 years ago
What are the physical properties of the sun?
Butoxors [25]
Sunspot cycles, solar flares, prominences, layers of the Sun, coronal mass<span> ejections, and </span>nuclear<span> reactions</span>
5 0
3 years ago
45 m = how many cm ?
svet-max [94.6K]

Answer:

The answer is 4,500 cm

Explanation:

3 0
4 years ago
Read 2 more answers
Please help I’ll make u brainliest please
Ira Lisetskai [31]

Answer:

14J

20000J

200J

Explanation:

formula :W=FS

W:work done

F:force (N)

S:displacement moved in direction of force (m)

1) 20N×0.7m

=14J

2) 400N×50m

= 20000J

3) 10N×20m

=200J

8 0
2 years ago
In a simple electric circuit, Ohm's law states that V=IRV=IR, where V is the voltage in volts, I is the current in amperes, and
VMariaS [17]

To solve this problem, apply the concepts given from Ohm's Law. From there we will obtain the derivative of the function with respect to time and with the previously given values we will proceed to find the change in current as a function of the derivative

V = IR

Here

I = Current

V = Voltage

R = Resistance

Taking the derivative we will have,

\frac{dV}{dt} = \frac{dI}{dt}R + I \frac{dR}{dt}

Our values are given as,

\frac{dV}{dt} = -0.01

\frac{dR}{dt} = 0.04\Omega/s

R = 300\Omega

I = 0.01A

Replacing we will have that

\frac{dV}{dt} = \frac{dI}{dt}R + I \frac{dR}{dt}

-0.01 = \frac{dI}{dt}(300\Omega) + (0.01A)(0.04\Omega/s)

Rearranging to find the current through the time,

\frac{dI}{dt} = \frac{-0.01- (0.01A)(0.04\Omega/s)}{(400\Omega)}

\frac{dI}{dt} = -0.000026A/s

Therefore the change of the current is -0.000026A per second

3 0
3 years ago
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