A roller coaster pushes a 25 kg person upward with a force of 300 N. What is the acceleration?

Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f

stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2$

When s = Db, t = 2t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2$

Dividing the two equations

$\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db$

Hence, Da=(1/4)Db

3 0

4 years ago

A swimming pool has the shape of a box with a base that measures 30 m by 12 m and a uniform depth of 2.2 m. How much work is req

Alex777 [14]

Answer:

853776 J

Explanation:

The work-energy needs to pump water out of the pool is the product of the weight of water and distance h

E = Wh = mgh

Since water mass is a body of water we can treat it as the product of density 1000kg/m3 and volume, which is the product of base area and uniform height h

$m = \rho V = \rho \int\limits^{2.2}_0 {A} \, dh$

Therefore:

$E = mgh = g\rho A\int\limits^{2.2}_0 {h} \, dh\\E = 9.8*1000*30*12[h^2/2]^{2.2}_0 = 1764000(2.2^2 - 0^2) = 853776 J$

5 0

3 years ago

Fan can accelerate from the starting blocks to running 11m/s down the track in 5 seconds. What is her acceleration?

Firdavs [7]

To answer this question we subtract the initial velocity from her final velocity. On the starting blocks the initial velocity was 0 m/s. Her final velocity was 11 m/s.t = 11m/s - 0 m/s, so the change in velocity is 11m/s. Time was 5 seconds.Plug it into the formula: 11 m/s ÷ 5 s= 2.2 m/s<span>2</span>

4 0

3 years ago

To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 130 m. The car

marysya [2.9K]

Answer:

0.739

Explanation:

If we treat the four tire as single body then

W ( weight of the tyre ) = mass × acceleration due to gravity (g)

the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r

where v is speed 25.6 m/s and r is the radius of the circle

centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²

net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²

coefficient of static friction between the tires and the road = frictional force / force of normal

frictional force = m × net acceleration / m×g

where force of normal = weight of the body in opposite direction

coefficient of static friction = (7.2567 × m) / (9.81 × m)

coefficient of static friction = 0.739

4 0

3 years ago

Under what conditions will an object experience a gravitational force?

Radda [10]

A) an object with mass > 0 in a gravitational field
b) an object with an electric charge not 0 in an electric field
c) a moving object with an electric charge not 0 in a magnetic field

6 0

3 years ago