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Art [367]
3 years ago
6

Ryan is driving his car to band practice. His speed is 55 mph. Splat! A bug smashes against the windshield, and bug "guts" are e

verywhere. Gross! Which scenario is true?
a. The windshield exerts a greater force on the bug than the bug exerts on the windshield.
b. The bug exerts a greater force on the windshield than the windshield exerts on the bug.
c. The force that the windshield exerts on the bug and the force that the bug exerts on the windshield are the same magnitude.
Physics
2 answers:
Zina [86]3 years ago
5 0

Answer:

B.

Explanation:

The bug is not travelling 55mph and is much smaller than the car

MrRa [10]3 years ago
3 0

Answer:

The answer is C.

Explanation:

Every point mass attracts every single other point mass by a force acting along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them. In magnitude, the force they apply each other is the same. Therefore, the force that the windshield exerts on the bug and the force that the bug exerts on the windshield are the same magnitude.

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URGENT! 40 POINTS! For each situation below, find the work done on the box using both energy and forces.
TiliK225 [7]

Answer:

i’m not sure one second let me try to figure it out

Explanation:

4 0
2 years ago
if the net force of an object is in a positive direction what will the direction of the resulting acceleration be
myrzilka [38]
Resulting acceleration would be in Positive direction, 'cause their direction doesn't change & they both have same direction.

Hope this helps!
3 0
3 years ago
What will be the final velocity of a 5.0 g bullet starting from rest, if a net force of 45 N is applied over a time of 0.02s?
Wewaii [24]

We know there’s a change in momentum due to a force applied over a time interval. Ft= m[v(final)-v(initial)]. Now simply plug in know values: (45)(0.02)=.005[v(final)-0]. Remember converting grams to kilograms. Solve for v final

3 0
3 years ago
A cylinder-piston system contains an ideal gas at a pressure of 1.5 105 pa.
Sedbober [7]

The change in the internal energy of the ideal gas is determined as -28 J.

<h3>Work done on the gas</h3>

The work done on the ideal gas is calculated as follows;

w = -PΔV

w = -1.5 x 10⁵(0.0006 - 0.0002)

w = -60 J

<h3>Change in the internal energy of the gas</h3>

ΔU = w + q

ΔU = -60J + 32 J

ΔU = -28 J

Thus, the change in the internal energy of the ideal gas is determined as -28 J.

Learn more about internal energy here: brainly.com/question/23876012

#SPJ1

5 0
2 years ago
RACTIC PTUDIES
Amanda [17]

Answer:

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

Explanation:

The distance of the book before the lamp is moved, d_{b} = 30 cm

The distance of the book after the lamp is moved, d_{a} = 90 cm

Illumination can be given by the formula, E = \frac{P}{4 \pi d^{2} }

Illumination before the lamp is moved, E_{b} = \frac{P}{4 \pi d_{b} ^{2} }

Illumination after the lamp is moved, E_{a} = \frac{P}{4 \pi d_{a} ^{2} }

\frac{E_{a}}{E_{b}} } = \frac{\frac{P}{4 \pi d_{a} ^{2} } }{\frac{P}{4 \pi d_{b} ^{2} } }

\frac{E_{a} }{E_{b} } = \frac{d_{b} ^{2} }{d_{a} ^{2}} \\\frac{E_{a} }{E_{b} } = \frac{30^{2} }{90 ^{2}}\\\frac{E_{a} }{E_{b} } =\frac{900 }{8100}\\\frac{E_{a} }{E_{b} } =\frac{1 }{9}

E_{b} = 9E_{a}

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

6 0
3 years ago
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