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3 years ago

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An object falls from a high building and hits the ground in 8.0 seconds. Ignoring air resistance, what is the distance that it f

ell?

1 answer:

d1i1m1o1n [39]3 years ago

4 0

Answer:

310 meters

Explanation:

Given:

v₀ = 0 m/s

t = 8.0 s

a = -9.8 m/s²

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (8.0 s) + ½ (-9.8 m/s²) (8.0 s)²

Δy = -313.6

Rounded to two significant figures, the object fell 310 meters.

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A student conducts an experiment in which a cart is pulled by a variable applied force during a 2sec time interval. In trial 1,

Harlamova29_29 [7]

Answer:

change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

Explanation:

Let's look for the speed of the car

F = m a

a = F / m

We use kinematics to find lips

v = v₀ + a t

v = v₀ + (F / m) t

The moment is defined by

p = m v

The moment change

Δp = m v - m v₀

Let's replace the speeds in this equation

Δp = m (v₀ + F / m t) - m v₀

Δp = m v₀ + F t - m v₀

Δp = F t

We see that the change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

8 0

3 years ago

The fulcrum of a first-class lever divides its 9.0 m arm into two sections—a 6.0 m arm and a 3.0 m arm. You place a rock weighin

nexus9112 [7]

For balancing the lever, force on both the sides shall be equal. so,
Force on 3 m end = m × a = 3 × 98.1 = 294.3

Now, on 6 m end, it would be: = 294.3/6 = 49.05
After rounding-off to the nearest hundredth value, it would be: 49 N

Finally, Option A would be your correct answer.

Hope this helps!

6 0

2 years ago

Which would have the longer orbital period: a moon 1 million km from the center of Jupiter, or a moon 1 million km from the cent

Harman [31]

Answer:

earth

Explanation:

The formula for the orbital period of the moon is given by

$T = 2\pi \sqrt{\frac{r}{g}}$

As the time period is inversely proportional to the square root of the acceleration due to gravity of the planet.

As the value of acceleration due to gravity on Jupiter is more than the earth, so the period of moon around the earth is large as compared to the period of the moon around the Jupiter when the distance is same.

5 0

3 years ago

A 5.0 kg chunk of putty moving at 10m/s collides and sticks to a 7.0 kg bowling ball that is initially at rest.What is the total

Flura [38]

Answer:

Total momentum = 50kgm/s

Explanation:

<u>Given the following data;</u>

Mass, M1 = 5kg

Mass, M2 = 7kg

Velocity, V1 = 10m/s

Velocity, V2 = 0m/s (since it's at rest).

To find the total momentum;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = Mass * Velocity$

The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.

Total momentum = M1V1 + M2V2

Substituting into the equation, we have;

Total momentum = 5*10 + 7*0

Total momentum = 50 + 0

<em>Total momentum = 50 kgm/s</em>

<em>Therefore, the total momentum of the bowling ball and the putty after they collide is 50 kgm/s. </em>

8 0

3 years ago

A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is th

Soloha48 [4]

Answer:

The velocity of mass 2m is $v_B = 0.67 m/s$

Explanation:

From the question w are told that

The mass of the billiard ball A is =m

The initial speed of the billiard ball A = $v_1$ =1 m/s

The mass of the billiard ball B is = 2 m

The initial speed of the billiard ball B = 0

Let the final speed of the billiard ball A = $v_A$

Let The finial speed of the billiard ball B = $v_B$

According to the law of conservation of Energy

$\frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Substituting values

$\frac{1}{2} m (1)^2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Multiplying through by $\frac{1}{2}m$

$1 =v_A^2 + 2 v_B ^2 ---(1)$

According to the law of conservation of Momentum

$mv_1 + 2m(0) = mv_A + 2m v_B$

Substituting values

$m(1) = mv_A + 2mv_B$

Multiplying through by $m$

$1 = v_A + 2v_B ---(2)$

making $v_A$ subject of the equation 2

$v_A = 1 - 2v_B$

Substituting this into equation 1

$(1 -2v_B)^2 + 2v_B^2 = 1$

$1 - 4v_B + 4v_B^2 + 2v_B^2 =1$

$6v_B^2 -4v_B +1 =1$

$6v_B^2 -4v_B =0$

Multiplying through by $\frac{1}{v_B}$

$6v_B -4 = 0$

$v_B = \frac{4}{6}$

$v_B = 0.67 m/s$

4 0

3 years ago