A car is accelerating at 30 m/s2, if the car is 400 kg how much force

1 answer:

6 0

It would be 12,000 because newton’s third 2nd law states F=ma (force=matter x acceleration) so 30x400 would be your force .

please mark brainliest and i hope this helps!

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A fire truck has a searchlight with a resistance of 60 (ohm) which is placed across a 24-V battery. What is the current in this

Mariulka [41]

Answer: 0.4 Amps

Explanation:

Voltage of battery = 24 Volts

Current I = ?

Resistance of searchlight (R)= 60ohms (Ω is the symbol for ohms)

Then, apply the formula for ohms law

Voltage = Current x resistance

i.e V = I x R

24V = I x 60Ω

I = 24V / 60Ω

I = 0.4 Amps (Amps is the unit of current)

Thus, the current in the circuit is 0.4 Amps

3 0

3 years ago

so I know you can solve this either by using Vox or Voy. I'm getting 3.08s when using Vox and 3.14s for Voy way. For Voy I'm usi

Tatiana [17]

The person's horizontal position is given by

$x=v_0\cos40^\circ t$

and the time it takes for him to travel 56.6 m is

$56.6\,\mathrm m=\left(24.0\,\dfrac{\mathrm m}{\mathrm s}\right)\cos40^\circ t\implies t=3.08\,\mathrm s$

so your first computed time is the correct one.

The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity $v_y$ at time $t$ is

$v_y=v_{0y}-gt$

which tells us that he would reach the ground at about $t=3.15\,\mathrm s$ . In this time, he would have traveled

$x=v_{0x}(3.15\,\mathrm s)=57.9\,\mathrm m$

But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity $v_{0y}$ would have been a bit smaller than $\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ$ .