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Aleonysh [2.5K]
3 years ago
11

7. Who provided evidence for the existence of an atomic nucleus?

Physics
1 answer:
andreyandreev [35.5K]3 years ago
4 0

Answer: Ernest Rutherford

Explanation:

Ernest Rutherford discovered the nucleus of the atom in 1911. He sent a beam of alpha particles toward gold foil and observed the way the particles were deflected by the gold atoms.

(please mark me brainliest)

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Scenario
zepelin [54]

Answer:

 t = 23.255 s,   x = 2298.98 m,    v_y = - 227.90 m / s

Explanation:

After reading your extensive writing, we are going to solve the approach.

The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.

As there is a mixture of units in different systems we are going to reduce everything to the SI system.

         v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s

         y₀ = 2650 m

Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement

Y axis  

       y = y₀ + v₀ t - ½ g t²

the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero

       0 = y₀ + 0 - ½ g t²

       t = \sqrt{  \frac{2 y_o}{g} }

       t = √(2 2650/ 9.8)

       t = 23.255 s

Therefore, for the cargo to reach the desired point, it must be launched from a distance of

       x = v₀ₓ t

       x = 111.76 23.255

       x = 2298.98 m

at the point and arrival the speed is

        vₓ = v₀ₓ = 111.76

     

vertical speed is

         v_y = v_{oy} - gt

          v_y = 0 - gt

          v_y = - 9.8 23.25 555

         v_y = - 227.90 m / s

the negative sign indicates that the speed is down

in the attachment we have a diagram of the movement

7 0
3 years ago
A racetrack has the shape of an inverted cone, as the drawing shows. On this surface the cars race in circles that are parallel
sergey [27]

Answer:

The value of d is 183.51 m.

Explanation:

Given that,

Speed of car = 34.0 m/s

Suppose The car race in the circle parallel to the ground surface is at an angle 40°

The radius of circular path r = d\cos\theta

Normal force acting on the car = N

We need to calculate the value of d

Using component of normal force

The horizontal component of normal force is equal to the gravitational force.

N\cos\theta=mg....(I)

The vertical component of normal force is equal to the centripetal force

N\sin\theta=\dfrac{mv^2}{r}.....(II)

Divided equation (I) by equation (II)

\tan\theta=\dfrac{v^2}{gr}

Put the value of g

\tan\theta=\dfrac{v^2}{g\times d\cos\theta}

v^2=\tan\theta\times g\times d\cos\theta

v^2=g\times d\sin\theta

d=\dfrac{v^2}{g\sin\theta}

Put the value into the formula

d=\dfrac{(34.0)^2}{9.8\times\sin40}

d=183.51\ m

Hence, The value of d is 183.51 m.

3 0
3 years ago
If a freely suspended vertical spring is pulled in downward direction and then released, which type of wave is produced in the s
larisa [96]

Answer:

longitudinal wave

Explanation:

it is perpendicular to the direction of the wave

3 0
3 years ago
Read 2 more answers
shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at const
Contact [7]

Answer:

a) T=0.40 N

b) T=1.9 s

Explanation:

Let's find the radius of the circumference first. We know that bob follows a circular path of circumference 0.94 m, it means that the perimeter is 0.94 m.

The perimeter of a circunference is:

P=2\pi r=0.94

r=\frac{0.94}{2\pi}=0.15 m

Now, we need to find the angle of the pendulum from vertical.

tan(\alpha)=\frac{r}{L}=\frac{0.15}{0.90}=0.17

\alpha=9.44 ^{\circ}

Let's apply Newton's second law to find the tension.

\sum F=ma_{c}=m\omega^{2}r

We use centripetal acceleration here, because we have a circular motion.

The vertical equation of motion will be:

Tcos(\alpha)=mg (1)

The horizontal equation of motion will be:

Tsin(\alpha)=m\omega^{2}r (2)

a) We can find T usinf the equation (1):

T=\frac {mg}{cos(\alpha)}=\frac{0.04*9.81}{cos(9.44)}=0.40 N

We can find the angular velocity (ω) from the equation (2):

\omega=\sqrt{\frac{Tsin(\alpha)}{mr}}=3.31 rad/s

b) We know that the period is T=2π/ω, therefore:

T=\frac{2\pi}{\omega}=\frac{2\pi}{3.31}=1.9 s

I hope it helps you!

8 0
2 years ago
A block is sent up a frictionless ramp along which an x axis extends upward. The figure below gives the kinetic energy of the bl
ss7ja [257]
Kinetic energy =1/2 mv^2 

<span>m=2ke/v^2 </span>

<span>m=2(34)/3.6^2 </span>

<span>m=5.24 </span>

<span>force normal = mg </span>
<span>=5.24 x 9.8 </span>
<span>force normal = 51.4N

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.


</span>
5 0
3 years ago
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