How much force is needed to accelerate a 66 kg skier at 2 m/sec^2?

A graduated beaker with 375 mL of water is sitting on a scale which measures the weight of the glass and water to be 7.60 N. Whe

Allisa [31]

Answer:

Volume of water displaced = 450 - 375 = 75 ml

Vr = volume of rock = 75 ml

Wr = 9.22 - 7.60 = 1.62 N weight of 75 ml of rock

Density of rock = 1.62 N / 75 ml = .0216 N / ml

Density of water = 1000 g / 1000 ml = 9.8 N / 1000 ml = .0098 N / ml

Density of rock / density of water = .0216 / .0098 = 2.20

8 0

2 years ago

Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.) With what initial v

lukranit [14]

Answer:

Vi = 94.64 m/s

Explanation:

I order to find out the initial velocity of the object, we can use third equation of motion:

2ah = Vf² - Vi²

where,

a = acceleration = -9.8 m/s²

h = maximum height covered by object = 460 m - 3 m = 457 m

Vf = Final Velocity = 0 m/s (since, object momentarily stops at highest point)

Vi = Initial Velocity = ?

Therefore,

2(-9.8 m/s²)(457 m) = (0 m/s)² - Vi²

Vi = √8957.2 m²/s²

Vi = 94.64 m/s

3 0

3 years ago

How do u do 5? Show me plz

nikitadnepr [17]

Solve the equation.it is hard to see

6 0

2 years ago

A hot-air balloon and its basket are accelerating upward at 0.265 m/s2, propelled by a net upward force of 688 N. A rope of negl

Sergeu [11.5K]

Question: Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)

Answer:

The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon $m_1$ and the mass of the basket $m_2$ , then according to newton's second law:

$(1). \:F = (m_1+m_2)a$ ,

where $a =0.265m/s^2$ is the upward acceleration, and $F = 688N$ is the net propelling force (counts the gravitational force).

Also, the tension $T$ in the rope is 79.8 N more than the basket's weight; therefore,

$(2). \:T = m_2g+79.8$

and this tension must equal

$T -m_2g =m_2a$

$(3). \:T = m_2g +m_2a$

Combining equations (2) and (3) we get:

$m_2a = 79.8$

since $a =0.265m/s^2$ , we have

$\boxed{m_2 = 301.13kg}$

Putting this into equation (1) and substituting the numerical values of $F$ and $a$ , we get:

$688N = (m_1+301.13kg)(0.265m/s^2)$

$\boxed{m_1 = 2295 kg}$

Thus, the mass of the balloon and the basket is 2295 kg and 301 kg respectively.

8 0

3 years ago

If a baseball pitch leaves the pitcher's hand horizontally at a velocity of 150 km/h by what percent will the pull of gravity ch

Slav-nsk [51]

0.52% First, let's convert that speed into m/s. 150 km/h * 1000 m/km / 3600 s/h = 41.667 m/s Now let's see how much time gravity has to work on the ball. Divide the distance by the speed. 18 m / 41.667 m/s = 0.431996544 s Now multiply that time by the gravitational acceleration to see what the vertical component to the ball's speed that gravity adds. 0.431996544 s * 9.8 m/s^2 = 4.233566131 m/s Use the pythagorean theorem to get the new velocity of the ball. sqrt(41.667^2 + 4.234^2) = 41.882 m/s Finally, let's see what the difference is (41.882 - 41.667)/41.667 = 0.005159959 = 0.5159959% Rounding to 2 figures, gives 0.52%

8 0

3 years ago