How far (in meters) will you travel in 3 minutes running at a rate of 6 m/s?

The magnetic field at the equator points north. if you throw a positively charged object (for example, a baseball with some elec

asambeis [7]

Recall the equation for magnetic force:

F = qv x B *x is cross product, not separate variable!

If the magnetic field points towards N and you throw E, then the magnetic force would point up, or out of the page. Use the right-hand rule. You point your finger towards the direction of the object, and curl your finger to the magnetic field. Your thumb is the direction of the magnetic force.

Hope this helps!

7 0

3 years ago

Two identical bowling balls are rolling on a horizontal floor without slipping. The initial speed of both balls is v = 10 m/s. B

Lapatulllka [165]

Answer:

The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.

We will use conversation of energy.

$K_A_1 + U_A_1 = K_A_2 + U_A_2\\\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_A$

Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.

For the ball B;

$K_B_1 + U_B_1 = K_B_2 + U_B_2$

$\frac{1}{2}I_B\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_B$

The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.

Hence, both balls climb the same point.

Explanation:

4 0

3 years ago

[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r

blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b) it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity $V_{f}$ for an object traveling distance h taking the initial vertical component of velocity as $V_{i}$ the kinematics equation is written as

$V_{f}^{2}=V_{i}^{2}+2ah$ where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

$V_{f}^{2}=V_{i}^{2}+2gh$

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

$V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}$ = 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

$V_{f}^{2}=V_{i}^{2}+2gh$

Since we have final velocity of 25 m/s

$V_{i}^{2}=2gh-V_{f}^{2}$

$V_{i}=\sqrt{(V_{f}^{2}-2gh)}$

$V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}$ = 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0

3 years ago

Javier wants to improve his flexibility and increase his personal discipline. Which community resource should he select?

Dovator [93]

Karate class, kicking high and ducking fast makes you much more flexible and you must pay respects to those ou fight with and be intune with yourself to do it right, in other words karate is the answer.

7 0

3 years ago

At what distance along the central perpendicular axis of a uniformly charged plastic disk of radius 0.390 m is the magnitude of

aliina [53]

<span>Radius, the distance from the centre = 0.390
Electric field is equal to half of the magnitude. E2 = E / 2
Given
E1 = E2 E1 = k x Q / r^2
E2 = (k x Q / r2^2) / 2
Equating the both we get 2 x r^2 = r2^2
r2 = square root of (2 x r1^2) = square root of (2) x r = 1.414 x 0.390
r2 = 1.414 x 0.390 = 0.551 m</span>

3 0

2 years ago