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irinina [24]
3 years ago
12

Help please only answer if you really know i need this ill give 50 points ;( thx

Physics
1 answer:
kakasveta [241]3 years ago
6 0
4, 1, 3, 2 ??

I think that’s right
You might be interested in
2. A force of 156 N acts on a 7.3-kg bowling ball for 0.40 s.
atroni [7]

Answer:

62.4N\cdot s

Explanation:

The change in momentum of the ball is equal to the impulse exerted on it, therefore:

\Delta p = I = F \Delta t

where

\Delta p is the change in momentum

F is the average force exerted on the ball

\Delta t is the time during which the force is applied

In this problem,

F = 156 N

\Delta t = 0.40 s

So, the change in momentum of the ball is

\Delta p =(156)(0.40)=62.4N\cdot s

6 0
4 years ago
The answer is 84N, but what is the mass and weight calculation?
Monica [59]

Answer:

weight= density ×volume× 9.8

mass=density×volume

Explanation:

weight = 0.12 \times 0.04 \times 0.04 \times 3.14 \times 13600 \\  \times 9.8 = 80.351 \: n \\  \\ mass = 0.12 \times 0.04 \times 0.4 \times 3.14 \times 13600 =  \\  = 8.199 \: kg

4 0
3 years ago
Your dog (who weighs 20kg) has just fallen off a cliff! Oh no!! Luckily they were on a leash, and you (who weighs 70kg) was fort
Maksim231197 [3]

Answer:

yes

Explanation:

becasue 70 divided by 0.3 = >20

Hope I helped:)

3 0
3 years ago
A large lightning bolt is observed to have a 19500 A current and move 36 C of charge. What was its duration?
Lynna [10]

Answer:

Its duration is 1.85*10⁻³ s or 1.85 ms

Explanation:

The intensity of electric current I is defined as the amount of electric charge Q (measured in Coulombs) that passes through a section of a conductor in each unit of time. The letter I is used to name the Intensity and its unit is the Ampere (A).

The intensity of electric current is expressed as:

I=\frac{Q}{t}

where:

I: Intensity expressed in Amps (A)

Q: Electric charge expressed in Coulombs (C)

t: Time expressed in seconds (s)

Being:

  • I= 19500 A
  • Q=36 C
  • t=?

Replacing:

19500 A=\frac{36 C}{t}

Solving:

19500 A*t= 36 C

t=\frac{36 C}{19500 A}

t= 1.85*10⁻³ s= 1.85 ms (being 1 s= 1,000 ms)

<u><em>Its duration is 1.85*10⁻³ s or 1.85 ms</em></u>

8 0
3 years ago
How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 4.00
klio [65]

Explanation:

Let us assume that the separation of plate be equal to d and the area of plates is 9 \times 10^{-4} m^{2}. As the capacitance of capacitor is given as follows.

            C = \frac{\epsilon_{o}A}{d}

It is known that the dielectric strength of air is as follows.

               E = 3 \times 10^{6} V/m

Expression for maximum potential difference is that the capacitor can with stand is as follows.

                       dV = E × d

And, maximum charge that can be placed on the capacitor is as follows.

               Q = CV

                   = \frac{\epsilon_{o} A}{d} \times E \times d

                   = \epsilon_{o}AE

                   = 8.85 \times 10^{-12} \times 3 \times 10^{6} \times 4 \times 10^{-4}

                   = 1.062 \times 10^{-8} C

or,                = 10.62 nC

Thus, we can conclude that charge on capacitor is 10.62 nC.

5 0
4 years ago
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