answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
DaniilM [7]
2 years ago
13

A ball is released from a tower at a height of 100 meters toward the roof of another tower that is 25 meters high. The horizonta

l distance between the two towers is 20 meters. With what horizontal velocity should the ball be imparted so that it lands on the rooftop of the second building?
Physics
1 answer:
Sindrei [870]2 years ago
3 0
Let's figure out how long it will take to fall 75 meters, from the 1st roof to the second roof. We can assume that gravity is the only force affecting vertical velocity, so the ball starts from rest and accelerates downward at g, or -9.8m/s².

Let's find how long it takes to fall 75 meters:
s(t) = Vi + (1/2)*a*t²,
where s(t) is displacement as a function of time, Vi is initial velocity (zero), and a is acceleration. Plugging in our values:
-75 = 0 + (1/2)(-9.8)(t²)      Multiply both sides by 2/-9.8
15.3 = t²                             Take the square root of both sides
t = 3.91

We need to the ball to travel 20 meters horizontally before it hits the roof in 3.91 seconds. We can assume that the horizontal velocity remains constant (a=0, Vi=V(t) for all t). 
Therefore, the minimum horizontal velocity is:
D = V*t , simple distance formula, distance equals velocity times time:
20 = V * 3.91       Divide both sides by 3.91
V = 5.11

The horizontal velocity, therefore, must be at least 5.11m/s in order for the ball to reach the roof of the second building. 
You might be interested in
Which occurrence would lead you to conclude that lights are connected in a
skelet666 [1.2K]

Answer:B When one bulb burns out, all the others lights stay lit.

Explanation:

3 0
2 years ago
A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a
o-na [289]

Answer:395.6 m/s

Explanation:

Given

mass of bullet m=5 gm

mass of wood block M=1 kg

Length of string L=2 m

Center of mass rises to an height of 0.38 cm

initial velocity of bullet u=450 m/s

let v_1 and v_2 be the velocity of bullet and block after collision

Conserving momentum

mu=mv_1+Mv_2 -------------1

Now after the collision block rises to an height of 0.38 cm

Conserving Energy for block

kinetic energy of block at bottom=Gain in Potential Energy

\frac{Mv_2^2}{2}=Mgh_{cm}

v_2=\sqrt{2gh_{cm}}

v_2=\sqrt{2\times 9.8\times 0.38}

v_2=0.272 m/s

substitute the value of v_2 in equation 1

5\times 450=5\times v_1+1000\times 0.272

v_1=395.6 m/s

4 0
2 years ago
In the 25-ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int
Ugo [173]

Answer:

a. 8.33 x 10 ⁻⁶ Pa

b. 8.19 x 10 ⁻¹¹ atm

c. 1.65 x 10 ⁻¹⁰ atm

d. 2.778 x 10 ⁻¹⁴ kg / m²

Explanation:

Given:

a.

I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s

P rad = I / us

P rad  = 2500 W / m² / 3.0 x 10 ⁸ m/s

P rad = 8.33 x 10 ⁻⁶ Pa

b.

P rad = 8.33 x 10 ⁻⁶ Pa *[  9.8 x 10 ⁻⁶ atm / 1 Pa ]

P rad = 8.19 x 10 ⁻¹¹ atm

c.

P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]

P rad = 1.67 x 10 ⁻⁵ Pa

P₁ = 1.013 x 10 ⁵ Pa /atm

P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm

d.

P rad  = I / us

ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²

ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²

3 0
2 years ago
A squirrel in a tree drops an acorn. how long does it take the acorn to fall 20 feet?
mart [117]

We use the equation of motion,

S= ut+\frac{1}{2}at^{2}

Here, S is the height, u is initial velocity and a is acceleration.

Given, S = 20 \ ft S = 20 \ ft = 20 \times\frac{1 \ m}{3.2808399 ft}  = 6.096 \ m

As  acorn falls from tree, therefore we take the value of a = 9.8 \ m/s^2 and initial velocity u = 0.

Substituting these values in equation of motion,

6.096 \ m = 0 \times t +\frac{1}{2} \times 9.8 m/s^2 (t)^2 \\\\\ t = 1.12 \ s

Thus, the time taken by the acorn to fall 20  feet ( 6.096 m ) is 1.12 s.

5 0
2 years ago
Neha and Suhani are playing with two identical pendulums. They leave the bob from a certain position and wait for it to return t
Anika [276]

Answer:

The correct option is C

Explanation:

The pendulum bob would return at the same time because the initial angle a pendulum bob is dropped does not affect it's period (the time it takes for the pendulum to move back and forth), however the one with a larger angle move faster but would eventually arrive at the same "starting point" due to varying displacements made.

6 0
2 years ago
Other questions:
  • During a car accident, a 125 kg driver is moving at 31 m/s and in 1.5 seconds is brought to rest by an inflating air bag. What i
    14·2 answers
  • two forces have the same magnitude F, what is the angle between the two vectors if their sum has a magnitude of (a) 2F? (b) sqrt
    12·1 answer
  • The acceleration of an object as a function of time is given by a(t) = (1.00 m/s2)t2. If displacement of the object between time
    7·1 answer
  • A uniform thin circular rubber band of mass M and spring constant k has an original radius R?
    11·1 answer
  • If you double the mass of an object while leaving the net force unchanged what is the result
    7·1 answer
  • Assume you are given an int variable named nElements and a 2-dimensional array that has been created and assigned to a2d. Write
    11·1 answer
  • A charge of 4.9 x 10-11 C is to be stored on each plate of a parallel-plate capacitor having an area of 150 mm2 and a plate sepa
    5·1 answer
  • a 2.0 kg hoop rolls without slipping on a horizontal surface so that its center proceeds to the right with a constant linear spe
    7·1 answer
  • Ben walks 500 meters from his house to the corner store. He then walks back toward his house, but continues 200 meters past his
    15·1 answer
  • What is Otter's average velocity over his entire trip when it takes him 2 minutes to walk 100 meters north and another 1 minute
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!