To verify the law of solid friction project work of physics

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?

Explanation:

The given data is as follows.

Mass of proton = $1.67 \times 10^{-27}$ kg

Charge of proton = $1.6 \times 10^{-19} C$

Speed of proton = $2.50 \times 10^{6} m/s$

Distance traveled = $5.31 \times 10^{-13} m$

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

$(K.E + P.E)_{initial}$ = $(K.E + P.E)_{final}$

$(\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)$

where, $\frac{kq_{p}q_{t}}{r} = U = Electric potential energy$

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

Putting the given values into the above formula as follows.

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

= $(\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})$

= $5.218 \times 10^{-15} J$

Therefore, we can conclude that the electric potential energy of the proton and nucleus is $5.218 \times 10^{-15} J$ .

4 0

4 years ago

Liz puts a 1 kg weight and a 10 kg on identical sleds. She then applies a 10N force to each sled. Describe why the smaller weigh

elena-14-01-66 [18.8K]

the correct answer is D- Acceleration depends indirectly on the mass.

According to Newton's second law, the force F, the mass m and the acceleration a are related as follows:

$F=ma$

Therefore,

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

The acceleration a₁ of the mass m₁ =1 kg is given by,

$a_1=\frac{F}{m_1} \\ =\frac{10N}{1kg} \\ =10m/s^2$

The acceleration a₂ of the mass m₂=10 kg is given by,

$a_2=\frac{10N}{10kg} \\ =1m/s^2$

The smaller mass has greater acceleration.

Thus, when the force applied on two bodies of different masses remains constant, then,

$a\alpha \frac{1}{m}$

Acceleration is inversely proportional to the mass of the body.

8 0

3 years ago

Read 2 more answers

2. If Peter is traveling .25 m/s on his bike, how long will it take him to reach

Semmy [17]

Answer:

48 seconds

Explanation:

Since S=d/t, plug in the known values and solve

0.25m/s=12m/t

0.25m/s*t=12m

t=12m/0.25m/s

t=48 seconds

7 0

3 years ago

Read 2 more answers

One ounce of a well-known breakfast cereal contains 104 Calories (1 food Calorie = 4186 J). If 2.5% of this energy could be conv

GalinKa [24]

Answer:

482.4 kg

Explanation:

104 calorie = 435344 J

The 2.5% of this energy equals to

435344 * 0.0025 = 10883.6 J

If this energy is converted to work done on lifting a barbell by a distance of 2.3. By the formula of Work W = force * distance then the force he exerts on the barbell is

$F = \frac{W}{D} = \frac{10883.6
}{2.3} = 4732 N$

Assuming gravitational constant g = 9.81 m/s2 then the mass of the barbell is

m = 4732 / 9.81 = 482.4 kg

5 0

3 years ago

A man stands on the roof of a 15.0-m-tall building and throws a rock with a speed of 30.0 m>s at an angle of 33.0%1b above th

Brilliant_brown [7]

The motion described here is a projectile motion which is characterized by an arc-shaped direction of motion. There are already derived equations for this type of motions as listed:

Hmax = v₀²sin²θ/2g
t = 2v₀sinθ/g
y = xtanθ + gx²/(2v₀²cos²θ)

where

Hmax = max. height reached by the object in a projectile motion
θ=angle of inclination
v₀= initial velocity
t = time of flight
x = horizontal range
y = vertical height

Part A.

Hmax = v₀²sin²θ/2g = (30²)(sin 33°)²/2(9.81)
Hmax = 13.61 m

Part B. In this part, we solve the velocity when it almost reaches the ground. Approximately, this is equal to y = 28.61 m and x = 31.91 m. In projectile motion, it is important to note that there are two component vectors of motion: the vertical and horizontal components. In the horizontal component, the motion is in constant speed or zero acceleration. On the other hand, the vertical component is acting under constant acceleration. So, we use the two equations of rectilinear motion:

y = v₀t + 1/2 at²
28.61 = 30(t) + 1/2 (9.81)(t²)
t = 0.839 seconds

a = (v₁-v₀)/t
9.81 = (v₁ - 30)/0.839
v₁ = 38.23 m/s

Part C.
y = xtanθ + gx²/(2v₀²cos²θ)
Hmax + 15 = xtanθ + gx²/(2v₀²cos²θ)
13.61 + 15 = xtan33° + (9.81)x²/[2(30)²(cos33°)²]
Solving using a scientific calculator,
x = 31.91 m

3 0

3 years ago

To verify the law of solid friction project work of physics ​

To verify the law of solid friction project work of physics