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Vlada [557]
2 years ago
10

Screw examples in daily life (simple machines)

Physics
1 answer:
postnew [5]2 years ago
5 0
Brace and bits, meat grinder, submarine hatch, screw lid jar, drills, machine screws, door lock, wooden handled cork screw. 
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An electron moving at 2.97×103 m/s in a 1.25 T magnetic field experiences a magnetic force of 1.40×10−16 N . What angle (in degr
lora16 [44]

To solve the problem it is necessary to take into account the concepts related to the Magnetic Force, which is given by,

F= qvBsin\theta

Where

F= Magnetic force

q= charge of proton

v= velocity

B Magnetic field

\theta=Angle between the velocity and the magnetic field.

Re-arrange the equation to find the angle we have,

\theta = sin^{-1}(\frac{F}{qvB})

Replacing our values we have,

V= 2.97*10^3m/s\\B = 1.25T\\F = 1.4*10^{-16}N\\q = 1.6*10^{-19}C\\

Then,

\theta = sin^{-1}(\frac{1.4*10^{-16}}{(1.6*10^{-19})(1.25)(2.97*10^3)})\\\theta = 13.63\°

The angle between 0 to 180 degrees would be,

\theta' = 180-13.63\\\theta' = 166.36

Therefore the two angles required are 13.63° and 166.36°

5 0
3 years ago
During a demolition derby, vehicle 1 collided vehicle 2 going in the opposite direction and came to an immediate stop. If the ve
snow_tiger [21]

Answer:

Their change in momentum is the same in magnitude and opposite in direction

Explanation:

The momentum of an object is defined as:

p=mv

where

m is the mass of the object

v is the velocity of the object

Therefore, the change in momentum of an object is

\Delta p = m\Delta v

where \Delta v is the change in velocity.

During a collision, the force experienced by an object is equal to the rate of change of momentum:

F=\frac{\Delta p}{\Delta t}

where \Delta t is the duration of the collision.

According to Newton's third law of motion, the force exerted by vehicle 1 on vehicle 2 during the collision is equal (and opposite) to the force exerted by vehicle 2 on vehicle 1, so

F_1=-F_2

Which means

\frac{\Delta p_1}{\Delta t}=-\frac{\Delta p_2}{\Delta t}

And since the duration of the collision is the same for the two vehicles, this becomes

\Delta p_1 =-\Delta p_2

7 0
3 years ago
A ball of mass 2 kg is moving with a speed of +6 m/s directly towards
shtirl [24]

Answer: 2.7

Explanation:

6 0
3 years ago
A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo
Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

W = \Delta U = U_2 - U_1

where the potential energy is defined as follows

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Let's first calculate the distance 'r' for both positions.

r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m

Now, we can calculate the potential energies for both positions.

U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J

Finally, the total work done on the moving particle can be calculated.

W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J

4 0
3 years ago
Read 2 more answers
A tennis player standing 12.6m from the net hits the ball at 3.00 degrees above the horizontal. To clear the net, the ball must
mezya [45]
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction. 
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end) 
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
4 0
3 years ago
Read 2 more answers
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