If you were to stir salt in water until the salt disappeared, which is the solute?

A hopper jumps straight up to a height of 1.3 m. With what velocity did he leave the floor

xxMikexx [17]

The velocity with which the jumper leaves the floor is 5.1 m/s.

<h3>What is the initial velocity of the jumper?</h3>

The initial velocity of the jumper or the velocity with which the jumper leaves the floor is calculated by applying the principle of conservation of energy as shown below.

Kinetic energy of the jumper at the floor = Potential energy of the jumper at the maximum height

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

v is the initial velocity of the jumper on the floor
h is the maximum height reached by the jumper
g is acceleration due to gravity

v = √(2 x 9.8 x 1.3)

v = 5.1 m/s

Learn more about initial velocity here: brainly.com/question/19365526
#SPJ1

3 0

9 months ago

Yea, gonna need some help. Thanks

natulia [17]

Answer:

t = 3.48 s

Explanation:

The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:

$h(t) = -16t^2+v_ot+h_o\\\\\frac{dh(t)}{dt}=0=-32t+v_o\\\\v_o = 32t$

where,

v₀ = initial speed = 110 ft/s

Therefore,

$110 = 32t\\\\t = \frac{110}{32}\\\\$

8 0

2 years ago

Sound travels through air at 343 m/s,

sasho [114]

The sound wave will have traveled 2565 m farther in water than in air.

Answer:

Explanation:

It is known that distance covered by any object is directly proportional to the velocity of the object and the time taken to cover that distance.

Distance = Velocity × Time.

So if time is kept constant, then the distance covered by a wave can vary depending on the velocity of the wave.

As we can see in the present case, the velocity of sound wave in air is 343 m/s. So in 2.25 s, the sound wave will be able to cover the distance as shown below.

Distance = 343 × 2.25 =771.75 m

And for the sound wave travelling in fresh water, the velocity is given as 1483 m/s. So in a time interval of 2.25 s, the distance can be determined as the product of velocity and time.

Distance = 1483×2.25=3337 m.

Since, the velocity of sound wave travelling in fresh water is greater than the sound wave travelling in air, the distance traveled by sound wave in fresh water will be greater.

Difference in distance covered in water and air = 3337-772 m = 2565 m

So the sound wave will have traveled 2565 m farther in water than in air.

5 0

3 years ago

How can a magnetic field be produced, using a wire, a battery, and and a nail?

Aleks [24]

It is fairly easy to build an electromagnet. All you need to do is wrap some insulated copper wire around an iron core. If you attach a battery to the wire, an electric current will begin to flow and the iron core will become magnetized. When the battery is disconnected, the iron core will lose its magnetism. Follow these steps.
Step 1 - Gather the Materials
One iron nail fifteen centimeters (6 in) long
Three meters (10 ft) of 22 gauge insulated, stranded copper wire
One or more D-cell batteries
Step 2 - Remove some Insulation
Step 3 - Wrap the Wire Around the Nail
Step 4 - Connect the Battery

5 0

3 years ago

Read 2 more answers

A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper

Levart [38]

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

∑ F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

cos 41 = T₁ₓ / T1

sin 41 = T₁y / T1

T₁ₓ = T₁ cos 41

T₁y= T₁ sin 41

for cable gold

cos 63 = T₂ / T₂

sin 63 = $T_{2y}$ / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

T₃ - W = 0

T₃ = W

T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

- T₁ₓ + T₂ₓ = 0

T₁ₓ = T₂ₓ

T1 cos 41 = T2 cos 63

T1 = T2 cos 63 / cos 41 (1)

y Axis

$T_{1y}$ + T_{2y} - T3 = 0

T₁ sin 41 + T₂ sin 63 = T₃ (2)

to solve the system we substitute equation 1 in 2

T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

T₂ (cos 63 tan 41 + sin 63) = W

T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

T₂ = 200 / (cos 63 tan 41 + sin 63)

T₂ = 200 / 1,2856

T₂ = 155.6 N

we substitute in 1

T₁ = T₂ cos 63 / cos 41

T₁ = 155.6 cos63 / cos 41

T₁ = 93.6 N

therefore the tension in each cable is

T₁ = 93.6 N

T₂ = 155.6 N

T₃ = 200 N

6 0

2 years ago