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3 years ago

2. (1) A piece of rubber is 50 cm long when a weight of

Physics

1 answer:

Harrizon [31]3 years ago

8 0

Answer

By F = -kx {-ve just indicating the sign of the force}

=>35 = k x (85-50) x 10^-2

=>k = 100 N/m

Again by F = -kx

You might be interested in

A table tennis ball has a mass of about 2.45 g. Suppose the ball is hit across the table with a speed of 4.0 m/s. What is it’s k

cluponka [151]

Answer:

0.0196 j

Explanation:

i) The formula for kinetic energy is as follows: 0.5*m*v^2

ii) Since we have all the values all that's left is to plug them into the equation

iii) First, WE MUST, Convert grams into kgs as this is the SI unit of mass so 2.45/1000

iv) All that's left now is to plug it into the equation so:

0.5* (s.45/1000)*(4^2)

v) Lastly we add the unit joules at the end as we're talking about energy

Hope this was useful! :)

3 0

2 years ago

Help please!!!

exis [7]

Answer:

5 years worth of work (aka all of the homework i currently have)

3 0

3 years ago

microwave ovens rotate at a rate of about 6.3 rev/min. what is this in revolutions per second? (you do not need to enter any uni

levacccp [35]

The microwave ovens rotate at a rate of about 0.105 rev/s.

The microwave rotation is the number of revolutions in a unit of time. To change the unit for angular velocity, assume that the quantity is multiplied by the unit it has. Then change to the desired units. The angular velocity is denoted by ω and has a magnitude of 6.3 rev/min.

ω = 6.3 rev/min

$\omega = 6.3 \times \frac{1 \: rev}{1 \: min}$

1 minute = 60 seconds
The revolution unit didn't change

$\omega = 6.3 \times \frac{1 \: rev}{60 \: seconds}$

$\omega = \frac{6.3 \: rev}{60 \: seconds}$

ω = 0.105 rev/s

Learn more about Angular velocity here: brainly.com/question/29344944

#SPJ4

7 0

1 year ago

In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere

muminat

Answer:

$2.1\cdot 10^{21}$ electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

$E=\frac{kQ}{r^2}$

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a charged sphere, so we have:

$E_b=3.00\cdot 10^6 N/C$ is the maximum electric field strength tolerated by the air before breakdown occurs

$r=1.00 km = 1000 m$ is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

$Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C$

Assuming that the cloud is negatively charged, then

$Q=-333.3 C$

And since the charge of one electron is

$e=-1.6\cdot 10^{-19}C$

The number of excess electrons on the cloud is

$N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}$

5 0

3 years ago

"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and

spayn [35]

Answer:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

8 0

3 years ago