Ask question

Ask question

All categories

harkovskaia [24]

2 years ago

7

In a shot-put competition, a shot moving at 15m/s has 450J of mechanical kinetic energy. What is the mass of the shot? Please he

lp, and include the formula for the answer and a step by step explanation

1 answer:

Llana [10]2 years ago

5 0

Answer:

Mass of shot (m) = 4 kg

Explanation:

Given:

Velocity (v) = 15 m/s

Mechanical kinetic energy (K.E) = 450 J

Find:

Mass of shot (m) = ?

Computation:

Mechanical kinetic energy (K.E) = 1/2mv²

Mechanical kinetic energy (K.E) = [1/2](m)(15)²

450 = [1/2](m)(15)²

900 = 225 m

Mass of shot (m) = 4 kg

You might be interested in

010) Identify the true statement. Group of answer choices The height of waves is determined by wind strength and fetch. Wave bas

AlekseyPX

Answer:

The height of the wave is determined by the wind strength and fetch.

Explanation:

The height of the wave is determined by the wind strength and fetch.

The more the strength and the more the fetch size the more will be the height of the wave.

Remember as the wave approaches the coast its wavelength decreases and the wave height increases, whereas when the wave goes away from the coast its wavelength increases and height decreases.

7 0

2 years ago

According to the author, Picasso’s artistic achievements were in large part the result of his:

Harrizon [31]

Answer:

Picasso’s artistic achievements were in large part the result of his contribution to help bring the Nazi' devastating casual bombing and Spanish civil war in Guernica to the world's attention through his paintings.

4 0

2 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

2 years ago

The coordinates of a bird flying in the xy-plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2.part a:Cal

8090 [49]

Α $=2.4 \frac{m}{s}$

β $=1.2 \frac{m}{s^2}$

$x(t)=at$

$y(t)=3-$ β $t^2$

$Vx(t)=$ α

$Vy(t)=-2$ β $t$

$vectorV=[$ α $;-2$ β $t]$

$ax(t)=0$

$ay(t)=-2$ β $t$

$vector a [0;-2$ β $t]$

3 0

2 years ago

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

SCORPION-xisa [38]

Answer:

The value is $r = 5.077 \ m$

Explanation:

From the question we are told that

The Coulomb constant is $k = 9.0 *10^{9} \ N\cdot m^2 /C^2$

The charge on the electron/proton is $e = 1.6*10^{-19} \ C$

The mass of proton $m_{proton} = 1.67*10^{-27} \ kg$

The mass of electron is $m_{electron } = 9.11 *10^{-31} \ kg$

Generally for the electron to be held up by the force gravity

Then

Electric force on the electron = The gravitational Force

i.e

$m_{electron} * g = \frac{ k * e^2 }{r^2 }$

$\frac{9*10^9 * (1.60 *10^{-19})^2 }{r^2 } = 9.11 *10^{-31 } * 9.81$

$r = \sqrt{25.78}$

$r = 5.077 \ m$

7 0

2 years ago