A baseball with a mass of 0.80 kg is given an acceleration of 20.00 m/s. How much net force was applied to the ball

A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi

grin007 [14]

Explanation:

It is given that,

Magnitude of charge, $q=15\ \mu C=15\times 10^{-6}\ C$

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, $B=0.08\ j$

Velocity, $v=(5\ cos25)i+(5\ sin25)j$

$v=[(4.53)i+(2.11)j]\ m/s$

We need to find the magnitude of force on the charge. Magnetic force is given by :

$F=q(v\times B)$

$F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]$

<em>Since</em>, $i\times j=k\ and\ j\times j=0$

$F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]$

$F=0.00000543\ kN$

$F=5.43\times 10^{-6}\ kN$

So, the force acting on the charge is $5.43\times 10^{-6}\ kN$ and is moving in positive z axis. Hence, this is the required solution.

6 0

3 years ago

What is the purpose of drawing a dohr model of an atom?

pav-90 [236]

To demonstrate the number of protons neutrons and electrons.

8 0

3 years ago

What is the final velocity, in meters per second, of a freight train that accelerates at a rate of 0.085 m/s2 for 7.5 min, start

Llana [10]

Answer:

$v_f=41.65\frac{m}{s}$

Explanation:

The final velocity is given by the following kinematic equation:

$v_f=v_0+at$

Here, $v_0$ is the initial velocity, a is the body's acceleration and t is the motion time. We have to convert the time to seconds:

$7.5min*\frac{60s}{1min}=450s$

Now, we calculate the final velocity:

$v_f=3.4\frac{m}{s}+(0.085\frac{m}{s^2}(450s))\\v_f=41.65\frac{m}{s}$

4 0

3 years ago

A 5.00X10^5 kg rocket is accelerating straight up. Its engines produce 1.250X10^7 N of thrust, and air resistance is 4.50X10^6 N

katrin [286]

<span>6.20 m/s^2 The rocket is being accelerated towards the earth by gravity which has a value of 9.8 m/s^2. Given the total mass of the rocket, the gravitational drag will be 9.8 m/s^2 * 5.00 x 10^5 kg = 4.9 x 10^6 kg m/s^2 = 4.9 x 10^6 N Add in the atmospheric drag and you get 4.90 x 10^6 N + 4.50 x 10^6 N = 9.4 x 10^6 N Now subtract that total drag from the thrust available. 1.250 x 10^7 - 9.4 x 10^6 = 12.50 x 10^6 - 9.4 x 10^6 = 3.10 x 10^6 N So we have an effective thrust of 3.10 x 10^6 N working against a mass of 5.00 x 10^5 kg. We also have N which is (kg m)/s^2 and kg. The unit we wish to end up with is m/s^2 so that indicates we need to divide the thrust by the mass. So 3.10 x 10^6 (kg m)/s^2 / 5.00 x 10^5 kg = 0.62 x 10^1 m/s^2 = 6.2 m/s^2 Since we have only 3 significant figures in our data, the answer is 6.20 m/s^2</span>

8 0

3 years ago

what sense can people use to detect infrared radiation? we can see it as red light, we can feel it as heat, we can hear it as a

Ad libitum [116K]

We can feel it as heat, using the nerve endings in our skins.

7 0

3 years ago

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