All categories

2 years ago

A few alpha particles bounce off a thin sheet of gold foil. What did scientists

Physics

1 answer:

Vilka [71]2 years ago

7 0

Answer:

I world say A is the answer 80% sure

You might be interested in

A car travels straight for 20 miles on a road that is 30Â° north of east. What is the east component of the carâ€™s displacement

Over [174]

The east component of the cars displacement is 17.3 miles.

Trigonometric ratio is used to show the relationship between the sides of a right angled triangle and its angles.

Let x represent the east component of the cars displacement.

Using trigonometric ratio:

cos(30) = x / 20

x = 20 * cos(30)

x = 17.3 miles

The east component of the cars displacement is 17.3 miles.

Find out more on Trigonometric ratio at: brainly.com/question/1201366

4 0

2 years ago

In noisy factory environments, it's possible to use a loudspeaker to cancel persistent low-frequency machine noise at the positi

Vaselesa [24]

Answer:

7.39 m or 3.61 m

Explanation:

$\lambda$ = Wavelength

f = Frequency = 90 Hz

v = Speed of sound = 340 m/s

Path difference of the two waves is given by

$s_1-s_2=\frac{\lambda}{2}$

Velocity of wave

$v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{340}{90}\\\Rightarrow \lambda=3.78\ m$

$s_1=s_2\pm\frac{\lambda}{2}\\\Rightarrow s_1=5.5\pm \frac{3.78}{2}\\\Rightarrow s_1=7.39\ m, 3.61\ m$

So, the location from the worker is 7.39 m or 3.61 m

7 0

3 years ago

Hey, earn points here

navik [9.2K]

Answer:yayyyy

Explanation:

4 0

3 years ago

Read 2 more answers

Calculate the molecular weight of Aluminium hydroxide

Vedmedyk [2.9K]

Al(OH)3 = 26.98 + [(16×3) + (1.01×3)] = 26.98 + 51.03 = 78.01 and the unit will be g/mol

5 0

3 years ago

If a basball is project upwards from the ground level with an initial velovaity of 32 feet per second, then it's height is a fun

inessss [21]

Answer:

Maximum height reached by the ball is 32 meters.

Explanation:

It is given that,

If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :

$s=-8t^2+32t$ ...........(1)

t is the time taken

s is the height attained as a function of time.

Maximum height achieved can be calculated as :

$\dfrac{ds}{dt}=0$

$\dfrac{d(-8t^2+32t)}{dt}=0$

-16 t + 32 = 0

t = 2 seconds

Put the value of t in equation (1) as :

$s=-8(2)^2+32(2)$

s = 32 meters

So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.

6 0

3 years ago