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3 years ago

According to Newton´s second law of motion, what does an object with more mass require? Question 8 options:

2 answers:

6 0

More force to accelerate. If you push a car compared to a beach ball, which will you have to push harder for it to move? That's pretty much what the question is asking, if that helps any:)

trapecia [35]3 years ago

6 0

more for force to accelerate

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What is kinetic energy

Elena L [17]

Answer:

energy which a body possesses by virtue of being in motion.

Explanation:

7 0

3 years ago

W=90 kg×4n/kg<br>Please halp me i need it

kiruha [24]

I'm guessing that this is a problem to find the weight of a 90kg mass on a planet where the acceleration of gravity is 4 m/s^2. (Much less gravity than Earth, a little more than Mars.)

Just do the multiplication, and you get

360 Newtons.

5 0

3 years ago

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$

To solve for "t" in this quadratic equation, we can factor it out as shown:

$0= t(20-\frac{9.8}{2} t)$

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

$0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s$

7 0

3 years ago

Which of the following statements is correct?

Dafna11 [192]

These are the correct solutions:

It is 11 a.m. in the Eastern Time Zone; therefore, it is 8 a.m. in the Pacific Time Zone. (3 hrs behind)

It is 3 p.m. in the Central Time Zone; therefore, 2 p.m. in the Mountain Time Zone. (1 hr behind)

It is 6 p.m. in the Pacific Time Zone; therefore, it is 4 p.m in Hawaii. (2 or 3 hours behind depending on time of year)

It is 6 p.m. in Hawaii; therefore, it is 11 p.m. in the Eastern Time Zone (5 or 6 hours behind depending on time of year).

It is 3 p.m. in Hawaii; therefore, it is 6 p.m. in the Mountain Time Zone (3 or 4 hours behind depending on time of year).

6 0

3 years ago

To a stationary, a bus moves south with a speed of 15 m/s. a man inside walks toward the front of the bus with a speed of 0.2 m/

marta [7]

It is 15.2 m/s South.

8 0

3 years ago

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