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3 years ago

Suri rolled a bowling ball down a lane in 2.5 s. The ball traveled

Physics

1 answer:

igor_vitrenko [27]3 years ago

7 0

Answer:

3.1 m/s

Explanation:

The acceleration of the ball is given by:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time elapsed

Here we know:

a = 1.8 m/s^2 is the acceleration

v = 7.6 m/s is the final velocity

t = 2.5 is the time elapsed

Solving for u, we find the initial velocity of the ball:

$u=v-at=7.6 m/s-(1.8 m/s^2)(2.5 s)=3.1 m/s$

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A dock worker loading crates on a ship finds that a 27 kg crate, initially at rest on a horizontal surface, requires a 80 N hori

Aleksandr-060686 [28]

Answer:0.302

Explanation:

Given

mass of crate m=27 kg

Force required to set crate in motion is 80 N

Once the crate is set in motion 56 N is require to move it with constant velocity

i.e. 80 N is the amount of force needed to just overcome static friction and 56 is the kinetic friction force

thus

$f_s(static\ friction)=\mu \cdot N$

where $\mu$ is the coefficient of static friction and N is Normal reaction

$N=mg$

$f_s=\mu mg$

$\mu mg=80$

$\mu =\frac{80}{27\times 9.8}$

$\mu =0.302$

7 0

3 years ago

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fredd [130]

Answer:

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3 years ago

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Rzqust [24]

Hi There,

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3 0

3 years ago

Read 2 more answers

An orbiting satellite can become charged by the photoelectric effect when sunlight ejects electrons from its outer surface. Sate

Rufina [12.5K]

Answer:

the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm

Explanation:

Given data

Φ = 5.32 eV

to find out

the longest wavelength

solution

we know that

hf = k(maximum) +Ф ...............1

here we consider k(maximum ) will be zero because photon wavelength max when low photon energy

so hf = 0

and hc/ λ = +Ф

so λ = hc/Ф ................2

now put value hc = 1240 ev nm and Φ = 5.32 eV

so hc = 1240 / 5.32

hc = 233 nm

the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm

8 0

3 years ago

011 10.0 points

Ulleksa [173]

Answer:

2.47 m

Explanation:

Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.

The horizontal velocity of the ball is constant:

$v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s$

and the time taken to cover the horizontal distance d is

$t=\frac{d}{v_x}=\frac{52}{20.3}=2.56 s$

So this is the time the ball takes to reach the horizontal position of the crossbar.

The vertical position of the ball at time t is given by

$y=u_y t - \frac{1}{2}gt^2$

where

$u_y = v sin \theta =(25)(sin 35.9^{\circ})=14.7 m/s$ is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:

$y=(14.7)(2.56) - \frac{1}{2}(9.8)(2.56)^2=5.52 m$

The height of the crossbar is h = 3.05 m, so the ball passes

$h' = 5.52- 3.05 = 2.47 m$

above the crossbar.

8 0

3 years ago