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zloy xaker [14]

3 years ago

11

A 0.400 kg bead slides on a straight frictionless wire and moves with a velocity of 3.50 cm/s to the right, as shown below. The

bead collides elastically with a larger 0.600 kg bead that is initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.70 cm/s.
What is the total kinetic energy of the system of beads after the collision?
A 1.40 × 10−4 J
B 2.45 × 10−4 J
C 4.70 × 10−4 J
D 4.90 × 10−4 J

1 answer:

irakobra [83]3 years ago

6 0

Answer:

B

Explanation:

KE = 1/2 m v^2 = 1/2 (0.400) (3.50^2)

= 2.45 J

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You drive a car 1600 ft to the east, then 2500 ft to the north. The trip took 2.5 minutes. What was the magnitude of your averag

elena-14-01-66 [18.8K]

Answer:

$Velocity=6.03m/s$

Explanation:

Given data

Time t=2.5 minutes=150 seconds

Distance A=1600 ft=487.68 m........east

Distance B=2500 ft=762m ........north

To find

Average velocity

Solution

First we need to find the resultant distance magnitude.To find that we apply Pythagorean theorem to find hypotenuse

So

$A^{2}+B^{2}=C^{2}\\ C=\sqrt{A^{2}+B^{2}}\\ C=\sqrt{(487.68m)^{2}+(762m)^{2}}\\ C=904.7m$

$Velocity=\frac{Distance}{Time}\\Velocity=\frac{904.7m}{150s}\\Velocity=6.03m/s$

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3 years ago

A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se

Mama L [17]

Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

$v = \frac{Bqr}{m}$

$v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}$

v = 661098.9 = 661099 m/s

(b)

$\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1\times 10^{-31}}$

e / m = 1.76 x 10^14 C / kg

(c) Let K be the kinetic energy

K = 0.5 x mv²

K = 0.5 x 9.1 x 10^-31 x 661099 x 661099

K = 1.99 x 10^-19 J

K = 1.24 eV

So, the potential difference is

V = 1.24 V

(d) if the acceleration voltage is doubled

V = 2 x 1.24 = 2.48 V

So, Kinetic energy

K = 2.48 eV

K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J

Let v is the speed

K = 0.5 x mv²

3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

v = 933856.5 m/s

Let the new radius is r.

$r=\frac{mv}{Bq}$

$r=\frac{9.1\times 10^{-31}\times 933856.5}{4.7\times 10^{-4}\times 1.6\times 10^{-19}}$

r = 0.0113 m = 1.13 cm

7 0

3 years ago

Imagine that the ball on the left is given a nonzero initial velocity in the horizontal direction, while the ball on the right c

masya89 [10]

Answer:

vₓ = xg/2y

Explanation:

In this question, let us find the time it takes for the ball on the right that has zero initial velocity to reach the ground.

By newton equation of motion we know that

y = v₀ t - ½ g t²

t = 2y / g

This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance

vₓ = x/t

vₓ = xg/2y

vₓ = xg/2y

Where we assume that x and y are known.

7 0

3 years ago

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

$KE =\frac{1}{2}mv^2$

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$2M_2V_1^2=0.5\times M_2V_2^2$

or

$2V_1^2=0.5\times V_2^2$

or

$4V_1^2= V_2^2$

or

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

or

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

or

$2(V_1+9.0)^2=(2V_1+9.0)^2$

or

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

or

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

or

$(0.404V_1)=(3.72)$

or

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

8 0

3 years ago

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Standing on the surface of the earth you have a weight of 100 N. If you were to travel until you were 2

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Answer:

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