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olganol [36]
3 years ago
7

Can somebody help please ! a. -8.3 m/s b.-4.2 m/s c.-0.12 m/s d. 0 m/s

Physics
1 answer:
Gnom [1K]3 years ago
4 0
The answer is a
the \: answer \: is \: a.
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1. How would you expect an instantaneous acceleration vs. time graph to look for a cart moving with a constant
r-ruslan [8.4K]

Answer:

a=0   v = v₀ + a t

a=0    line is horizontal

Explanation:

1, In a graph of acceleration vs. time, we have lines, when the line is horizontal it is zero, when the line has a positive slope the increasing accelerations and when the slope is negative the decreasing acceleration

2, speed and relationship of a car is given by

        v = v₀ + a t

where vo is the initial velocity, a is the acceleration and tel time

in this case I will calcograph velocity vs. time the constant acceleration is a straight line.

In general from the graph we can find the initial velocity with the cut at that x and the acceleration of the car with the slope

6 0
3 years ago
Helppp!!! what's the answer to this??
riadik2000 [5.3K]

Answer:

(b)

Explanation:

The voltage always lags the current by 90°, regardless of the frequency.

4 0
2 years ago
PLS HELP
Kazeer [188]
The answer would be letter choice B
7 0
3 years ago
Prove for N/C=Volt/m
Zigmanuir [339]

Answer:

simple, Volt =change in potential energy/Charge

the unit of energy is newton meter (Force*distance)

the unit of charge is coloumb

So, Volt/meter=newton* meter/coloumb*meter

=newton/coloumb (hence proved)

This unit is the potential drop per unit of length in a conductive wire with uniform resistance

7 0
3 years ago
A cannon fires a shell straight upward; 2.3 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming
PtichkaEL [24]

Answer:

The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.                    

Explanation:

Let u is the initial speed of the launch. Using first equation of motion as :

u=v-at

a=-g

u=v+gt\\\\u=17+9.8\times 2.3\\\\u=39.54\ m/s

The velocity of the shell at launch and 5.4 s after the launch is given by :

v=u-gt\\\\v=39.54-9.8\times 5.4\\\\v=-13.38\ m/s

So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

6 0
3 years ago
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