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Elina [12.6K]
3 years ago
14

A car has a mass of 1000 kg and accelerates at 2 meters per second per second. what is the magnitude of the net force exerted on

the car
Physics
2 answers:
9966 [12]3 years ago
4 0
F=ma
Therefore the net force = 1000kg × 2 metres per second per second
So F=2000 N
olga_2 [115]3 years ago
3 0

Answer:

The magnitude is 2000N

Explanation:

To solve this question we are going to use the second Newton law.

The second Newton law states that the acceleration of an object is directly proportional to the net force which acts on the object and inverse proportional to the mass of the object ⇒

a=\frac{F}{m}

Which can also be written as :

F=m.a (I)

Where F is the force

Where m is the mass of the object and where ''a'' is the acceleration.

In the problem, we know that

m=1000kg

a=2\frac{m}{s^{2}}

If we replace in the equation (I) :

F=(1000kg).(2\frac{m}{s^{2}})=2000(kg.\frac{m}{s^{2}})

The unit kg.\frac{m}{s^{2}} is defined as N which is called Newton. N is an unit of force.

Finally, the net force exerted on the car is 2000N

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Answer:

The volume of 3200 kg of petrol is 4 m^3.

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A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found
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Answer:

1)  \rm q_2 is<u> positive.</u>

<u></u>

2) \rm q_2=4.56\times 10^{-10}\ C.

Explanation:

<h2><u>Part 1:</u></h2>

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., \rm q_2 is <u>positive.</u>

<u></u>

<h2><u>Part 2:</u></h2>

<u></u>

<u>Given:</u>

  • Mass of the balloon, m = 0.00275 kg.
  • Charge on the balloon, \rm q_1 = -3.50\times 10^{-8}\ C.
  • Distance between the rod and the balloon, d = 0.0640 m.
  • Acceleration due to gravity, \rm g = 9.81\ m/s^2.

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, \rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.

The magnitude of the electrostatic force on the balloon due to the rod is given by

\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.

\rm \dfrac{1}{4\pi \epsilon_o} is the Coulomb's constant.

For the elecric force and the weight to be balanced,

\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.

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Which of these components is present in this circuit schematic?
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Its resistor :}...............

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What are the three most abundant gases in the dry atmosphere?
Anastaziya [24]

Answer:

Nitrogen, Oxygen, Argon.

Explanation:

The three (3) most abundant gases in the dry atmosphere are"

- Nitrogen

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These are not the only components of dry air. Dry atmosphere is made up of:

- 78.09% Nitrogen;

- 20.95% Oxygen;

- 0.93% Argon;

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6 0
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An electron in a cathode-ray beam passes between 2.5cm long parallel-plate electrodes that are 6.0mm apart. A 2.1mT, 2.5-cm-wide
Dmitry_Shevchenko [17]

Answer:

(a). The speed of electron is 1.56\times10^{7}\ m/s.

(b). The radius of electron is 4.2\ cm

Explanation:

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Length = 2.5 cm

Distance = 6.0 mm

Magnetic field = 2.1 T

Potential difference = 700 V

(a). We need to calculate the electron's speed

Using formula of speed

v=\sqrt{\dfrac{2eV}{m}}

Put the value into the formula

v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times700}{9.1\times10^{-31}}}

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v=1.56\times10^{7}\ m/s

(b). We need to calculate the radius of electron

Using formula of centripetal force

\dfrac{mv^2}{r}=qvB

r=\dfrac{mv}{qB}

Where,

m = mass of electron

v = speed of electron

r = radius

q = charge of electron

B = magnetic field

Put the value into the formula

r=\dfrac{9.1\times10^{-31}\times1.56\times10^{7}}{1.6\times10^{-19}\times2.1\times10^{-3}}

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(b). The radius of electron is 4.2 cm

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