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3 years ago

The first qstn plsss

Physics

1 answer:

-BARSIC- [3]3 years ago

6 0

answer:

the first one

Explanation:

i think it is this one because 4k is bigger than 2k . also it is asking which one has the lager magnitude so it is the first one

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Heat always travels from

Monica [59]

Answer:

from hot toward cold. Heat moves naturally by any of three means. The processes are known as conduction, convection, and radiation. Sometimes more than one may occur at the same time.

Explanation:

4 0

3 years ago

A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37° above horizontal]. It gets blocked just after re

brilliants [131]

1) Data:

Vo = 20 m/s
α = 37°
Yo = 0
Y = 3m

2) Questions: V at Y = 3m and X at Y = 3 m

3) Calculate components of the initial velocity

Vox = Vo * cos(37°) = 15.97 m/s

Voy = Vo * sin(37°) = 12.04 m/s

4) Formulas

Vx = constant = 15.97 m/s

X = Vx * t

Vy = Voy - g*t

Y = Yo + Voy * t - g (t^2) / 2

5) Calculate t when Y = 3m (first time)

Use g ≈ 9.8 m/s^2

3 = 12.04 * t - 4.9 t^2

=> 4.9 t^2 - 12.04t + 3 = 0

Use the quadratic equation to solve the equation

=> t = 0.28 s and t = 2.18s

First time => t = 0.28 s.

6) Calculate Vy when t = 0.28 s

Vy = 12.04 m/s - 9.8 * 0.28s = 9.3 m/s

7) Calculate V:

V = √ [ (Vx)^2 + (Vy)^2 ] = √[ (15.97m/s)^2 + (9.30 m/s)^2 ] = 18.48 m/s

tan(β) = Vy/Vx = 9.30 / 15.97 ≈ 0.582 => β ≈ arctan(0.582) ≈ 30°

Answer: V ≈ 18.5 m/s, with angle ≈ 30°

8) Calculate X at t = 0.28s

X = Vx * t = 15.97 m/s * 0.28s = 4,47m ≈ 4,5m

Answer: X ≈ 4,5 m

4 0

3 years ago

A 0.145 kg baseball is thrown with a velocity of 25.0 m/s. How much work was done on the baseball to bring it from rest to 25.0

sergey [27]

Answer:

45.31 J

Explanation:

We are given that

Mass of baseball , m=0.145 kg

Initial velocity, u=0

Final velocity, v=25 m/s

We have to find the work done on the baseball to bring it from rest to 25 m/s

We know that

Work done = Change in kinetic energy

Work done, W= $\frac{1}{2}m(v^2-u^2)$

Using the formula

Work done, W $=\frac{1}{2}(0.145)((25)^2-0)$

Work done= $\frac{1}{2}(0.145)(625)$

Work done, W=45.31 J

Hence, the work done on the baseball to bring it from rest to 25 m/s

=45.31 J

3 0

3 years ago

*example of previous incorrect/correct answer on similar question) Inside a vacuum tube, an electron is in the presence of a uni

andrezito [222]

The final speed of the electron is 4.64 * 10^5 m/s.

<h3>What is the speed of the electron?</h3>

Given that the mass of the electron is obtained as 9.11 * 10^-31 Kg, we have that the charge of the electron is 1.6 * 10^-19 C.

E= F/q

F = Eq

F = 330 N/C * 1.6 * 10^-19 C

F = 5.28 * 10^-17 N

F = ma

a = F/m = 5.28 * 10^-17 N/ 9.11 * 10^-31 Kg

a = 5.8 * 10^13 m/s^2

Using

v = u + at

u = 0 m/s because the electron was initially at rest

v = at

v = 5.8 * 10^13 * 8.00 ✕ 10−⁹ s

v = 4.64 * 10^5 m/s

Learn more about the speed of the electron:brainly.com/question/13130380

#SPJ1

8 0

2 years ago

A merry-go-round with a a radius of R = 1.99 m and moment of inertia I = 194 kg-m2 is spinning with an initial angular speed of

Aleksandr [31]

Answer:

Part 1)

$L_1 = 185.2 kg m^2/s^2$

Part 2)

$L_2 = 663.07 kg m^2/s^2$

Part 3)

$L = 663.07 kg m^2/s^2$

Part 4)

$\omega = 1.83 rad/s$

Part 5)

$F_c = 453.6 N$

Explanation:

Part a)

Initial angular momentum of the merry go round is given as

$L_1 = I \omega$

here we know that

$I = 194 kg m^2$

$\omega = 1.47 rad/s^2$

now we have

$L_1 = 194 \times 1.47$

$L_1 = 185.2 kg m^2/s^2$

Part b)

Angular momentum of the person is given as

$L = mvR$

so we have

$m = 68 kg$

$v = 4.9 m/s$

R = 1.99 m

so we have

$L_2 = (68)(4.9)(1.99)$

$L_2 = 663.07 kg m^2/s^2$

Part 3)

Angular momentum of the person is always constant with respect to the axis of disc

so it is given as

$L = 663.07 kg m^2/s^2$

Part 4)

By angular momentum conservation of the system we will have

$L_1 + L_2 = (I_1 + I_2)\omega$

$185.2 + 663.07 = (194 + 68(1.99^2))\omega$

$848.27 = 463.28 \omega$

$\omega = 1.83 rad/s$

Part 5)

Force required to hold the person is centripetal force which act towards the center

so we will have

$F_c = m\omega^2 R$

$F_c = 68(1.83^2)(1.99)$

$F_c = 453.6 N$

3 0

3 years ago

The first qstn plsss​

The first qstn plsss