You have a 6.00 V power supply, and

Greg is boiling water on the stove for some noodles. what object can he safely hold while draining the water from the noodles af

zmey [24]

In order for Greg to safely drain the water out of the noodles, he should use potholders or any thing that is does not conduct heat or transfer heat. Some pots are also equipped with handles that are made of plastics for safely transferring of its content to another container.

5 0

3 years ago

Can an ordinary object, like a motorcycle, be mass-less? Yes or No

Drupady [299]

Answer:

no.

Explanation:

because the mass of an object never changes.

4 0

3 years ago

How do you calculate the mass of an object accelerating at22.35m/s2 with a force of 120N

Arte-miy333 [17]

Answer:

The mass of object is calculated as 5.36 kg

Explanation:

The known terms to find the mass are:

acceleration of object (a) = 22.35 $m/s^{2}$

Force exerted (F) = 120N

mass of an object (m) = ?

From Newton's second law of motion;

F = ma

or, 120 = m × 22.35

or, m= $\frac{120}{22.35}$ kg

∴ m = 5.36 kg

3 0

3 years ago

The two conducting rails in the drawing are tilted upwards so they each make an angle of 30.0° with respect to the ground. The v

dolphi86 [110]

Answer:

The current flows through the rod is 14.9 A.

Explanation:

Given that,

Magnetic field = 0.045 T

Mass of aluminum rod = 0.19 kg

Length = 1.6 m

Angle = 30.0°

We need to calculate the force

Using resolving force

$F\cos\theta=mg\sin\theta$

$F=mg\tan\theta$

Put the value into the formula

$F=0.19\times9.8\times\tan30$

$F=1.075\ N$

We need to calculate the current flows through the rod

Using formula of magnetic force

$F=iLB$

$i=\dfrac{F}{LB}$

Put the value into the formula

$i=\dfrac{1.075}{1.6\times0.045}$

$i=14.9\ A$

Hence, The current flows through the rod is 14.9 A.

8 0

3 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

a)

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

b)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

c)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is: