Mass = m

1. A 1.30 kg ball strikes a wall with a velocity of -10.5 m/s. The ball bounces off with a velocity of 6.50 m/s. If the ball is

bekas [8.4K]

Let F be the magnitude of the force. The impulse of this force while the ball is in contact with the wall is

Ft = F (0.0210 s)

and this impulse is equal to the change in the ball's momentum,

m ∆v = (1.30 kg) (6.50 m/s - (-10.5 m/s)) = (1.30 kg) (17.0 m/s)

Solve for F :

F (0.0210 s) = (1.30 kg) (17.0 m/s)

F = (1.30 kg) (17.0 m/s) / (0.0210 s)

F ≈ 1050 N

4 0

3 years ago

In the figure, a proton is projected horizontally midway between two parallel plates that are separated by 0.50 cm, and are 5.60

noname [10]

a) Minimum speed of the proton: $6.05\cdot 10^6 m/s$

b) Angle of the velocity: $\theta=-5.1^{\circ}$

Explanation:

a)

The proton experiences a vertical force due to the electric field, given by:

$F=qE$

where

$q=1.6\cdot 10^{-19}C$ is the proton charge

$E=610,000 N/C$ is the magnitude of the electric field

The vertical acceleration of the proton is therefore

$a=\frac{qE}{m}$

where

$m=1.67\cdot 10^{-27}kg$ is its mass

Therefore, the vertical position of the proton at time t is

$y(t)=\frac{1}{2}at^2=\frac{1}{2}\frac{qE}{m}t^2$

where we assumed that the initial vertical velocity is zero (because the proton is fired horizontally) and the initial vertical position, halfway between the two plates, is the origin.

The horizontal motion of the proton instead is uniform, so the horizontal position is given by

$x(t)=v_0 t$

where $v_0$ is the initial speed. This equation can be rewritten as

$t=\frac{x(t)}{v_0}$

And substituting into the eq. for y,

$y(t) = \frac{1}{2}\frac{qE}{m} \frac{x^2}{v_0^2}$

Solving for the initial speed,

$v_0 = \sqrt{\frac{qEx^2}{2my}}$

The proton just misses one of the plate when

x = 5.60 cm = 0.056 m (length of the plates)

y = 0.25 cm = 0.0025 m (half the distance between the plates)

Therefore, we find the initial speed:

$v=\sqrt{\frac{(1.6\cdot 10^{-19})(610,000)(0.056)^2}{2(1.67\cdot 10^{-27})(0.0025)}}=6.05\cdot 10^6 m/s$

b)

In order to find the angle, we just need to analyze the horizontal and vertical component of the final velocity of the proton.

The horizontal velocity is constant so it is:

$v_x = v_0 = 6.05\cdot 10^6 m/s$

The vertical velocity is given by:

$v_y^2 - u_y^2 = 2ay$

where:

$u_y=0$ (initial vertical velocity is zero)

$a=\frac{qE}{m}$ (acceleration)

y = 0.0025 m (vertical displacement)

Solving for $v_y$ ,

$v_y = \sqrt{2ay}=\sqrt{2\frac{qEy}{m}}=5.4\cdot 10^5 m/s$

Therefore, the final angle of the velocity with respect to the horizontal is:

$\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.4\cdot 10^5}{6.05\cdot 10^6})=5.1^{\circ}$

And since the electric field is downward (the proton just misses the lower field), it means that the angle is below the horizontal:

$\theta=-5.1^{\circ}$

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

8 0

3 years ago

Which of the following items has the most inertia while at rest

Hunter-Best [27]

Answer:

wats the following items

3 0

3 years ago

A coil 3.55 cm in radius, containing 470 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.2

slavikrds [6]

The Electric current is 1.11* 10^{-4}A

Given that the coil's radius is 3.55 cm (0.35 m),

The formula for the coil's area is A = r2 A = (3.14) (0.35)2 = 0.005024 m2.

R = Resistance = 600 N = Number of spins = 500 B = Magnetic field = (0.0120)

t + (3 x 10⁻⁵) t⁴

The number t = 5 is substituted for taking the derivative at both the induced current and the electric current.

The Electric current is therefore 1.11* 10^{-4}A
Electric current - The rate of electron passage in a conductor is known as electric current. The ampere is the electric current's SI unit. Electrons are little particles that are part of a substance's molecular structure. These electrons can be held loosely or securely depending on the situation.

To learn more about electric current please visit -brainly.com/question/12791045
#SPJ1

7 0

2 years ago

Read 2 more answers

When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th

Veseljchak [2.6K]

Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant

The work done in stretching the rubber band from x = 0 to x = L is
$W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}$

Answer: $\frac{aL^{2}}{2}$

4 0

4 years ago