Each circuit has an equivalent power source, and all the resistors are

A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a right turn and travels 1.33 km east before maki

charle [14.2K]

Answer:

$AD\approx1.7582\ km$

Explanation:

Follow the schematic in which point A is the warehouse and point D is the destination.

Now we observe the triangle constructed $\Delta ADB$ :

here:

$AB\perp BD$

$AB=2.6-1.45$

$AB=1.15\ km$

&

$BD=1.33\ km$

As we know that displacement is the shortest distance between two points.

<u>Using Pythagoras theorem:</u>

$AD=\sqrt{AB^2+BD^2}$

$AD=\sqrt{(1.15)^2+(1.33)^2}$

$AD\approx1.7582\ km$

3 0

3 years ago

What is the significance of the fact that the gravitational constant, G, is a very small number and that Coulomb’s constant, k,

WITCHER [35]

The comparison indicates that the gravitational force is a much weaker force than the electromagnetic force. This is the significance of the fact that the gravitational constant, G, is a very small number and that Coulomb’s constant, k, is a very large number

Answer: The comparison indicates that the gravitational force is a much weaker force than the electromagnetic force.

<u>Explanation: </u>

We know that the universal formula for the estimation of gravitational force is,

$F=G \times\left(\frac{m_{1} \times m_{2}}{r^{2}}\right)$

Where,

F, Gravitational force, $m_{1} \text { and } m_{2}$ - objects masses r- The relative distance between the two objects

G, Gravitational Constant $6.67 \times 10^{-11} \frac{N m^{2}}{k g^{2}}$

And, the electrostatic force between two scalar charges according to the Coulomb’s law is,

$F=k \times\left(\frac{q_{1} \times q_{2}}{r^{2}}\right)$

F, Electrostatic force between two charges, $q_{1} \text { and } q_{2}$ - Two scalar charges, r - The relative distance between two scalar charges, k- Coulomb’s constant $8.987 \times 10^{9} \frac{N m^{2}}{c^{2}}$

Now, on comparing the values of G and k, we can easily evaluate that eventually, the gravitational force will be lesser than the coulomb’s force.

Besides this, we can also judge this fact through various examples such as, a balloon rubbed with a cloth, easily sticks to the wall for some time. Opposing the gravitational force of the Earth which is not the case with the normal balloon.

It drops without having the electrostatic force between the wall and the balloon. This shows that the gravitational force draws lesser impact on objects as compared to the electrostatic force.

7 0

3 years ago

Read 2 more answers

A student is late and running to class. A hall monitor tells them to slow down, so the student slows to a walking speed of 1.4 m

sergey [27]

Answer:

8.9m/s

Explanation:

Final velocity = initial velocity + acceleration *time

Vf=vi+at

Vf=1.4 m/s

Vi=?

A=-2.5m/s^2

T=3

1.4m/s=vi+(-2.5m/s^2)(3s)

Vi=8.9m/s

7 0

3 years ago

A cannon of mass 5.71 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7

zheka24 [161]

Answer:

541.14 m/s

Explanation:

We are given that

Mass of cannon= $m_1=5.71\times 10^3 kg$

Mass of shell, $m_2=73.5 kg$

Initial velocity of shell,v=547 m/s

We have to find the velocity of shell fired from this loose cannon.

According to law of conservation of momentum

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

Initial momentum of system=0

$m_1v_1=-m_2v_2$

$v_1=-\frac{m_2v_2}{m_1}$

When the cannon is bolted to the ground then only shell moves and kinetic energy of system equals to kinetic energy of shell

Kinetic energy of shell, $K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 73.5(547)^2}=1.09\times 10^7 J$

K.E of shell= $\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2$

K.E of shell= $\frac{1}{2}m_1(-\frac{m_2v_2}{m_1})^2+\frac{1}{2}m_2v^2_2$

K.E of shell= $\frac{1}{2}(\frac{m^2_2v^2_2}{m_1}+\frac{1}{2}m_2v^2_2)$

2K.E of shell= $m_2v^2_2(\frac{m_2}{m_1}+1)$

Velocity of shell fired from this loose cannon,v_2= $\sqrt{\frac{2k.E}{m_2(\frac{m_2}{m_1}+1)}$

$v_2=\sqrt{\frac{2\times 1.09\times 10^7}{73.5(\frac{73.5}{5.71\times 10^3}+1)}$

$v_2=541.14m/s$

Hence, the velocity of shell fired from this loose cannon would be 541.14 m/s

5 0

3 years ago

Mer and count Assessment items Assessment started:

RoseWind [281]

Answer:

Asteroids

Explanation:

Asteroids are the small and large block of rocks that are found to be scattering space. There are numerous asteroids present in the asteroid belt which is located between the planet of Mars and Jupiter. These range from a few meters to hundreds of kilometers. They move so rapidly, at about 80,000 km/hr and there occurs constant collision with one another. These asteroids sometimes enter into the atmosphere of earth, which can cause a great mass extinction event.

These are believed to be the remnants particles that are present in space right after the formation of the solar system.

5 0

4 years ago