Ask question

Ask question

All categories

alexandr402 [8]

3 years ago

14

A ball is thrown downward from the top of a 55.0 m tower with an initial speed of 11.0 m/s. Assuming negligible air resistance,

what is the speed of the ball just before hitting the ground?

1 answer:

irina1246 [14]3 years ago

8 0

I'm not positive however my guess is this.
KE=(1/2)mv^2
KE=(1/2)(55)(121)
KE=3327.5m/s^2
You might not want to use this tho.

You might be interested in

A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r

Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

(c) at = 4 m/s² , v = 42 m/s

Explanation:

The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t Formula (1)

θ= ω₀*t + (1/2)*α*t² Formula (2)

at = α*R Formula (3)

v= ω*R Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s : wheel’s initial angular velocity

R = 1.0 m : wheel’s radium

(a) Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ= ω₀*t + (1/2)*α*t²

θ= (2)*(10) + (1/2)*(4)*(10)²

θ= 220 rad

θ= 220 rad

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

6 0

3 years ago

a bicycle uniformly from rest at time t the velocity of the bicycle is v at what time will the bicycle have a velocity of 4v

sesenic [268]

Here

Acceleration and initial velocities are constant.

According to first equation of kinematics.

$\\ \sf\longmapsto v=u+at$

$\\ \sf\longmapsto v=0+at$

$\\ \sf\longmapsto v=at$

$\\ \sf\longmapsto v\propto t$

Time was t at velocity v
Time will be 4t at velocity 4v

7 0

3 years ago

A person suffering from anaemia gets tired after a short walk

FrozenT [24]

Answer:

yes

Explanation:

you will feel weary after shorter times

5 0

4 years ago

Read 2 more answers

An astronomer at the equator measures the Doppler shift of sunlight at sunset. From this, she calculates that Earth's tangential

Ksivusya [100]

Answer : Radius of Earth is 6,340 Km.

Explanation

It is given that,

Tangential velocity of Earth at the equator, v = 465 m/s

Centripetal acceleration at the equation is, $a_c=3.41\times 10^{-2}\ m/s^2$

We know that the relation between centripetal acceleration and tangential velocity is :

$a_c=\dfrac{v^2}{r}$ ..........(1)

Where,

v is tangential velocity

r is the radius

Putting all values in equation (1)

$3.41\times 10^{-2}\ m/s^2=\dfrac{(465 m/s)^2}{r}$

$r=6340909\ m$

or

$r=6,340\ Km$

The Earth's radius is 6,340 Km. Hence, this is the required solution.

7 0

3 years ago

A horizontal force of 14.0N is applied to a box of m=32.5kg with Vo=0. Ignoring friction, how far does the crate travel in 10.0s

Alex Ar [27]

I’m going to assume initial velocity is 0.

Use Newton’s second law:

F = m•a

F/m = a

14.0/32.5kg= 28/65 m/s^2

Use constant SUVAT acceleration formulae:

S- displacement - what we need to find out

U - initial velocity - 0

V

A - 28/65 m/s^2

T - 10 seconds

S = ut + 1/2at^2

Since u = 0

S = 1/2at^2

1/2• 28/65 • 10^2 = 21.5metres~

Answer is 21.5 metres

~Hoodini, here to help.

6 0

3 years ago