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3 years ago

What activity would require more impulse?

Physics

1 answer:

Gnesinka [82]3 years ago

3 0

B. Accelerating a bowling ball from rest to 35 m/s

Explanation:

Accelerating a bowling ball from rest to 35m/s will require more impulse compared to a baseball.

Impulse is the force acting on a body a particular period of time. It is similar to momentum.

When impulse is applied on a body, it change it state from rest and cause motion.

A body with more mass will require a higher impulse to cause it to accelerate. Bowling balls are heavier. They require more impulse to make them move.

Learn more:

Momentum brainly.com/question/9484203

#learnwithBrainly

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What is the change in velocity of a 22-kg object that experiences a force of 15 N for

vagabundo [1.1K]

Answer:

Force = mass × acceleration

Acceleration:

${ \tt{15 = (22 \times a)}} \\ { \tt{a = \frac{15}{22} \: {ms}^{ - 2} }}$

From first Newton's equation of motion:

${ \bf{v = u + at}}$

Change = v - u:

${ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \: {ms}^{ - 2} }}$

3 0

3 years ago

An asteroid orbits the Sun every 176 years. What is the asteroids average distance from the Sun? P ^ 2 = a ^ 3 where p = period

KatRina [158]

Answer:

The value is $x = 45.99 \ Au$

Explanation:

From the question we are told that

The period of the asteroid is $T = 176 \ years = 176 * 365 * 24 * 60* 60 = 5.55*10^{9}\ s$

Generally the average distance of the asteroid from the sun is mathematically represented as

$R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }$

Here M is the mass of the sun with a value

$M = 1.99*10^{30} \ kg$

G is the gravitational constant with value $G = 6.67 *10^{-11} \ m^3 \cdot kg^{-1} \cdot s^{-2}$

$R = \sqrt[3]{ \frac{6.67 *10^{-11} * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }$

=> $R = 6.88 *10^{12} \ m$

Generally

$1.496* 10^{11} \ m \to 1 Au (Astronomical \ unit )$

$R = 6.88 *10^{12} \ m \ \ \ \ \to \ \ x \ Au$

=> $x = \frac{6.88 *10^{12}}{1.496 *10^{11}}$

=> $x = 45.99 \ Au$

7 0

2 years ago

Help!! <br> A. 0.5 s <br> B. 1.0 s<br> C. 1.5 s<br> D. 2.0 s

algol13

Answer:

c. 1.5 s

Explanation:

i had this and i got it right

3 0

3 years ago

Um caminhão de entregas transporta produtos de tamanho e peso elevados, o que requer o uso de máquina simples para facilitar a d

Korvikt [17]

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8 0

3 years ago

Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of

Lana71 [14]

Electric field due to a point charge is given as

$E = \frac{kq}{r^2}$

here we know that

$q = 1.6 \times 10^{-19} C$

also the distance is given as

$r = 5.29 \times 10^{-11} m$

now we will have

$E = \frac{(9\times 10^9)(1.6 \times 10^{-19})}{(5.29 \times 10^{-11})^2}$

so we will have

$E = 5.14 \times 10^{11} N/C$

so above is the electric field due to proton

5 0

3 years ago