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aev [14]
4 years ago
3

The spring constant, k, for a 22cm spring is 50N/m. A force is used to stretch the spring and when it is measured again it is 32

cm long. Work out the size of this force.the
Physics
1 answer:
Romashka [77]4 years ago
8 0
1) Convert cm in m
2) Elastic energy formula:
U=1/2kΔx²
U=1/2×50×(0.32-0.22)
U=25×0.1
U=2.5 J
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Three charges 1.5*10-6, 3*10-6, -3*10-6 are placed at three vertices of an equilateral triangle of side 30cm. Find the net force
gulaghasi [49]

Answer:

F = 0N

Explanation:

The force between two charges is given by

F=k\frac{q_1q_2}{r^2}

where r is the distance between the charges and K is the Coulomb's constant

(k=8-89*10^9Nm^2/C^2)

The force in the first charge is only the sum of the forces due to the other charges. Hence we have

F_T=F_1+F_2=k\frac{q_2q_1}{r^2}+k\frac{q_3q_1}{r^2}

F_T=(8.89*10^9\frac{Nm^2}{C^2})\frac{(3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}+(8.89*10^9\frac{Nm^2}{C^2})\frac{(-3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}\\\\F_T=0.445N-0.445N=0N

Ft=0N

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5 0
3 years ago
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The mass of Tesla Model S is 2000 kg and its engine produces
yulyashka [42]

Answer:

The ratio of the power generated by the engines are \dfrac{2}{3}

(b) is correct option.

Explanation:

Given that,

Mass of Tesla = 2000 Kg

Torque = 400 Nm

Mass of Mercedes = 1.5 M Tesla

The Benz engine produces 50% more torque than the Tesla.

We need to calculate the power generated by the engines 6 seconds after a start from rest

Using formula of  power

P=F\cdot v

We know that,

v=at

v=\dfrac{F}{m}t

v=\dfrac{\tau}{mr}t

Put the value of F and v in equation (I)

P=\dfrac{\tau}{r}\times\dfrac{\tau}{mr}t

P=\dfrac{\tau^2t}{mr^2}

Here, r and t are same for both bodies.

P\propto\dfrac{\tau^2}{m}

We need to calculate the ratio of the power generated by the engines

Using formula of power

\dfrac{P_{T}}{P_{B}}=\dfrac{\tau_{T}^2}{m_{T}}\times\dfrac{m_{B}}{\tau_{B}}

Put the value into the formula

\dfrac{P_{T}}{P_{B}}=\dfrac{400^2}{2000}\times\dfrac{3000}{600^2}

\dfrac{P_{T}}{P_{B}}=\dfrac{2}{3}

Hence, The ratio of the power generated by the engines are \dfrac{2}{3}

3 0
3 years ago
) Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) . What are (a)
natali 33 [55]

To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

The vector product between the Force and the radius allows us to obtain the torque, in this way,

\tau = \vec{F} \times \vec{r}

\tau = (8i+6j)\times(-3i+4j)

\tau = (8*4)(i\times j)+(6*-3)(j\times i)

\tau = 32k +18k

\tau = 50 k

Therefore the torque on the particle about the origen is 50k

PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,

cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}

cos\theta = \frac{(8*-3)+(4*3)}{\sqrt{(-3)^2+4^2}*\sqrt{8^2+6^2}}

cos\theta = -0.24

\theta = cos^{-1} (-0.24)

\theta = 103.88\°

Therefore the angle between the ratio and the force is 103.88°

5 0
3 years ago
Because quantum mechanics is physics that describes the interactions of very small objects (i.e. molecules, atoms, and electrons
stira [4]
Ha! Lot of words but the question itself is easy.
The answer is 2.5 times 10 to the 5th power.
The main part of the numbers has the decimal point placed after the first digit.
Then for what number of power, you just count the number of decimal places moved.
I hope this helps you.
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3 years ago
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A virtual image produced by a lens is always
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C. located in front of the lens
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