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2 years ago

What is the KE of a 4 kg mass at 3 m/s?

Physics

2 answers:

AysviL [449]2 years ago

7 0

$\tt KE=\dfrac{1}{2}mv^2\\\\KE=\dfrac{1}{2}\times 4\times3^2\\\\KE=18~J$

Arlecino [84]2 years ago

4 0

Answer:

Explanation:

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A satellite’s velocity is 30000m/s. After 60 secs, it’s velocity shows to 15000m/a. What is the satellite’s acceleration?

Orlov [11]

Answer:

Acceleration = 9 × 10^5 m/s^2 ( deceleration )

Explanation:

From the first equation of motion:

V = u + at

15000 = 30000 + 60a

a = ( 15000-30000)/60

a = 9 × 10^5 m/s^2

4 0

2 years ago

Jin knows that the initial internal energy of a closed system is 78 J and the final internal energy is 180 J. He also knows that

lawyer [7]

As we know by the first law of thermodynamics

$Q = \Delta U + W$

here we know that

Q = heat given to the system

$\Delta U = U_f - U_i$

W = work done by the system

now here we can say

$\Delta U = 180 - 78 = 102 J$

$W = 64 J$

now we can say that heat will be given as

$Q = 64 + 102 = 166 J$

now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy

So best describe Jin's Error is

B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy.

8 0

3 years ago

Jennifer, who has a mass of 50.0 kg, is riding at 35.0 m/s in her red sports car when she must suddenly slam on the brakes to av

Assoli18 [71]

Answer:

Average force = 3.5 kN

Explanation:

Given:

Mass of Jennifer (m) = 50 kg

Initial velocity = 35 m/s

Time taken to stop body = 0.5 s

Find:

Average force

Computation:

v = u + at

0 = 35 + a(0.5)

Acceleration (a) = - 70 m/s² = 70 m/s²

Average force = ma

Average force = (50(70)

Average force = 3500 N

Average force = 3.5 kN

6 0

3 years ago

A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, ca

viva [34]

Answer:

$a=0.2*10^{-5}g$

Explanation:

From the question we are told that:

Mass $M=180=>0.18kg$

Charge $Q=18mC=18*10^-^3C$

Velocity $v=2.2m/s$

Length of Wire $L=8.6cm=>0.086$

Current $I=30A$

Generally the equation for Magnetic Field of Wire B is mathematically given by

$B=\frac{\mu_0*I}{2\pi*l}$

$B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}$

$B=6.978*10^{-5}T$

Generally the equation for Force on the plane F is mathematically given by

$F=qvB$

Therefore

$ma=qvB$

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

$a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}$

$a=2.37*10^{-5}$

Therefore in Terms of g's

$a=\frac{2.37*10^{-5}}{9.8}$

$a=0.2*10^{-5}g$

8 0

2 years ago

A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is

Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with uniform circular motion, her centripetal acceleration $a_{c}$ is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$a_{c}=17 m/s^{2}$ is the centripetal acceleration

$V$ is the tangential speed

$r=12 m$ is the radius of the circle

Isolating $V$ from (1):

$V=\sqrt{a_{c}r}$ (2)

$V=\sqrt{(17 m/s^{2})(12 m)}$

Finally:

$V=14.28 m/s$ This is the girl's tangential speed

3 0

3 years ago