Ask question

Ask question

All categories

OverLord2011 [107]

3 years ago

10

A water balloon was dropped from a high window and struck its target 1.1 seconds later. If the balloon left the person's hand at

-5 meters/sec what was the velocity on impact?

2 answers:

valkas [14]3 years ago

8 0

-5.5 is the answer to the question

Galina-37 [17]3 years ago

6 0

-5.5 is the correct answer for ur question

hope this helps u

You might be interested in

Suppose a thin conducting wire connects two conducting spheres. A negatively charged rod is brought near one of the spheres, the

dangina [55]

Answer:

a. The spheres will attract each other.

Explanation:

When two conducting spheres are connected by a conducting wire and a negatively charged rod is brought near it then this will induce opposite (positive) charge at the nearest point on the sphere and by the conservation of charges there will also be equal amount of negative charge on the farthest end of this conducting system this is called induced polarization.

When the conducting wire which joins them is cut while the charged rod is still in proximity to of one of the metallic sphere then there will be physical separation of the two equal and unlike charges on the spheres which will not get any path to flow back and neutralize.

Hence the two spheres will experience some amount of electrostatic force between them.

4 0

3 years ago

How would it be possible for an object to be traveling with a constant speed and still accelerating?

lys-0071 [83]

Yeah!! It's possible for an object by changing it's direction....

4 0

3 years ago

It would be really helpful if u help me solving this question. PLEASE!!!

sweet [91]

Answer: The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

The room temperature will be the point at which the themometric substance remains constant when rising from ice point.

Explanation:

Apparatus available:

Unmarked thermometer

250 cm3 glass beaker

crushed ice

water

heatproof mat

clamp, boss and stand

meter rule

Added apparatus

Bunsen burner

Stirrer

Method

The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

Procedures

Set up the apparatus of illustrated in the attached figure.

Immerse the unmarked thermometer into the ice in the beaker.

When the level indicated by the thermometric substance remains steady after some time, a mark will be made at that point. This mark will corresponds to the ice point (lower fixed point) and is assigned the value of 0 °C.

You may add little water and continue to stir gently.

The themometric substance will start to rise and stop when it reaches room temperature. Mark the point but do not assign any value

Place the beaker on bunsen burner and boil the water. The themometric substance will continue to rise and remain constant at upper fixed point

This mark will corresponds to the steam point (upper fixed point) and is assigned the value of 100 °C.

Divide between the lower fixed point and upper fixed point into equal intervals. Then you can see the value of room temperature.

7 0

3 years ago

Are dimensionless quantities always unitless

timofeeve [1]

<em>Important thing is that all unitless quantity is dimensionless quantity. .</em><em>A</em><em> dimensionless physical quantity may have an unit</em>

4 0

3 years ago

Read 2 more answers

A quarterback is set up to throw the football to a receiver who is running with a constant velocity v⃗ rv→rv_r_vec directly away

Artist 52 [7]

Answer:

a) V_o,y = 0.5*g*t_c

b) V_o,x = D/t_c - v_r

c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

d) Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

Explanation:

Given:

- The velocity of quarterback before the throw = v_r

- The initial distance of receiver = r

- The final distance of receiver = D

- The time taken to catch the throw = t_c

- x(0) = y(0) = 0

Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it. Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball. Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.

Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

x = x(0) + V_i,x*t_c

D = 0 + V_i,x*t_c

V_o,x + v_r = D/t_c

V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

V_o = sqrt ( V_o,x^2 + V_o,y^2)

V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

- The angle with which it should be thrown can be evaluated by trigonometric relation:

tan(Q) = ( V_o,y / V_o,x )

tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

6 0

3 years ago