Strong nuclear forces act over _____ distances within atoms.

A race car accelerates uniformly at 14.2 m/s2. If the race car starts from rest how fast will it

Kaylis [27]

Answer:

vf=94.4 m/s

Explanation:

acceleration is the final velocity minus initial velocity divided by time

a = (vf-vi)/t

Given:

a= 14.2 m/s^2

vi= 0 (at rest)

t = 6.6

Solve for vf

a = (vf-vi)/t

a*t=vf-vi

(14.2)*(6.6)=vf - 0

vf=94.4 m/s

6 0

3 years ago

Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k

Phantasy [73]

Answer:

Part(A): The magnitude of Jill's final velocity is $\bf{6.59~m/s}$ .

Part(B): The direction is $\bf{42.7^{0}}$ south to east.

Explanation:

Given:

Mass of Jack, $m_{1} = 59.0~Kg$

Mass of Jill, $m_{2} = 47..0~Kg$

Initial velocity of Jack, $v_{1i} = 8.00~m/s$

Initial velocity of Jill, $v_{2i} = 0$

Final velocity of Jack, $v_{1f} 5.00~m/s$

The final angle made by Jack after collision, $\alpha = 34.0^{0}$

Consider that the final velocity of Jill be $v_{2f}$ and it makes an angle of $\beta$ with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

$~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}$

where, $v_{2f}^{x}$ is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

$v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s$

Conservation of momentum of the system along north direction is given by

$~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s$

where, $v_{2f}^{y}$ is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

$v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s$

Part(B):

The direction is given by

$\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}$

4 0

3 years ago

Two resistors, R1 = 25.0 ohms and

Leni [432]

Answer: 0.078

Explanation:

Given variables:

R1 = 25.0 , R2 = 64.0 , I1 = 0.200

Step 1: find V

0.200 = v/25

0.200 * 25 = v

5 = v

Step 2: Solve for I2

I = V/R

I = 5/64

I = 0.078125

Hope this helps!

3 0

3 years ago

The mass of a cm3(cube) stone is 180.

ss7ja [257]

Density = (mass) / (volume), no matter how large or small the sample is.

We can't calculate the density, because you left out the number for the volume.
Also, you didn't tell us the unit for the mass of 180.

a). If the mass is 180 grams, then the density is

(180 gm) / (volume) .

b). No matter how many pieces you crush it into, and
no matter how large or small a piece is, its density is
the same. (I just wish we knew what the density really is.)

c). A piece may have 80 grams of mass. It doesn't "weigh" 80 grams.

Since the density of the whole rock is (180 gm) / (volume),
the density of any piece of it is (180 gm) / (volume).

Multiply each side by (volume): (Density) x (volume) = 180 gm

Divide each side by (density): Volume = (180 gm) / (density)

We can't calculate the volume of an 80-gm piece, because
we don't know the density. (That's because you left the volume
out of the question.)

4 0

3 years ago

The wires connecting the switches, light-bulbs and buzzers must start and end at the power source before the circuit can work. T

MakcuM [25]

A IS THE FINAL ANSWER :)

4 0

3 years ago

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