Answer:
Since the ball is moving by uniformly accelerated motion, its vertical velocity at time t is given by

where we took upward as positive direction, and where is the initial velocity, a the acceleration and t the time.

The instant at which is the instant when the ball reverses its velocity (from upward to downward). This means that the difference between the time t at which v(t)=0 and the instant t=0 is the total time during which the ball was going upward:

By plugging numbers into the equation, we find

where we took upward as positive direction, and where is the initial velocity, a the acceleration and t the time.

The instant at which is the instant when the ball reverses its velocity (from upward to downward). This means that the difference between the time t at which v(t)=0 and the instant t=0 is the total time during which the ball was going upward:

By plugging numbers into the equation, we find

If you have two substances, one with a density of 2.0 g/cm3 and one with a density of 1.3 g/cm3 and you combined them, which one would float on topother and why?

Consider position [x] = L, time [t] = T, velocity [v] = L/T and acceleration [a] = L/T 2 . Find the exponent A in the equation v = a^2 t^ A /x

A horizontal cylindrical tank 8.00 ft in diameter is half full of oil (60.0 Ib/ft3). Find the force on one end

Which of the following is correct? *PLEASE HELP MEEEE1 cm = 100 m1 mm = 100 cm100 mm = 1 cm1 m = 100 cm

A large convex lens stands on the floor. The lens is 180 cm tall, so the principal axis is 90 cm above the floor. A student holds a flashlight 120 cm off the ground, shining straight ahead (parallel to the floor) and passing through the lens. The light is bent and intersects the principal axis 60 cm behind the lens. Then the student moves the flashlight 30 cm higher (now 150 cm off the ground), also shining straight ahead through the lens. How far away from the lens will the light intersect the principal axis now?A. 30 cmB. 60 cmC. 75 cmD. 90 cm

Consider position [x] = L, time [t] = T, velocity [v] = L/T and acceleration [a] = L/T 2 . Find the exponent A in the equation v = a^2 t^ A /x

A horizontal cylindrical tank 8.00 ft in diameter is half full of oil (60.0 Ib/ft3). Find the force on one end

Which of the following is correct? *PLEASE HELP MEEEE1 cm = 100 m1 mm = 100 cm100 mm = 1 cm1 m = 100 cm

A large convex lens stands on the floor. The lens is 180 cm tall, so the principal axis is 90 cm above the floor. A student holds a flashlight 120 cm off the ground, shining straight ahead (parallel to the floor) and passing through the lens. The light is bent and intersects the principal axis 60 cm behind the lens. Then the student moves the flashlight 30 cm higher (now 150 cm off the ground), also shining straight ahead through the lens. How far away from the lens will the light intersect the principal axis now?A. 30 cmB. 60 cmC. 75 cmD. 90 cm

**Answer:**

Work done will be equal to 3186.396 J

**Explanation: **

We have mass m = 76.2 kg

Initial velocity u = 5 m/sec

Final velocity v = 10.4 m/sec

We have to find the work done

From work energy theorem work done is equal to change in kinetic energy

w = 3168.396 J

So work done will be equal to 3186.396 J

**Answer:**

it A

**Explanation:**

Its a negative ion that hss one less valence electron than a netural bromine atom

**Answer:**

**Explanation:**

Given that,

Vector A points in the -x direction with a magnitude of 21.

Let the x component is making an angle of 60 degrees with negative x axis. The x component of a vector is given by :

A = -42 units

The y component of a vector is given by :

**So, the y component of vector A is (-36.37) degrees. Hence, this is the required solution.**

If air were a good conductor of heat" then soup will not stay hot for longer because this time convection+conduction will both help to transfer heat away from soup. Because conduction is the transfer of heat through a substances as a result of neighbouring vibrating particles, The particles in air are far apart.

the answer is **false. **Hope this helps

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

The launching speed of the beetle is __ 6.4 m/s__.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

The time taken by the beetle to launch itself upwards is __ 1.62 ms__.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

The beetle can jump to a height of **2.1 m**

We have that for the Question the **Speed,T**ime and __Height are__

**u=6.38m/s****T=13sec****h=2m**

From the question we are told

- Certain
**insects**can achieve seemingly impossible accelerations**while**jumping.

- the click beetle
**accelerates**at an*astonishing*400g over a distance of 0.52 cm as it rapidly bends its*thorax*, making the "**click**" that gives it its name.

a)

Generally the equation for the *average ***velocity **is mathematically given as

**u=6.38m/s**

b)

Generally the equation for the **Time **of *flight*** **is mathematically given as

**T=13sec**

c)

Generally the equation for the *air ***resistance **is mathematically given as

**h=2m**

For more information on this visit

**Answer:**

Δ = 84 Ω, = (40 ± 8) 10¹ Ω

**Explanation:**

The formula for parallel equivalent resistance is

1 / = ∑ 1 / Ri

In our case we use a resistance of each

R₁ = 500 ± 50 Ω

R₂ = 2000 ± 5%

This percentage equals

0.05 = ΔR₂ / R₂

ΔR₂ = 0.05 R₂

ΔR₂ = 0.05 2000 = 100 Ω

We write the resistance

R₂ = 2000 ± 100 Ω

We apply the initial formula

1 / = 1 / R₁ + 1 / R₂

1 / = 1/500 + 1/2000 = 0.0025

= 400 Ω

Let's look for the error (uncertainly) of Re

= R₁R₂ / (R₁ + R₂)

R’= R₁ + R₂

= R₁R₂ / R’

Let's look for the uncertainty of this equation

Δ / = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

ΔR’= ΔR₁ + ΔR₂

We substitute the values

Δ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

Δ / 400 = 0.1 + 0.05 + 0.06

Δ = 0.21 400

Δ = 84 Ω

Let's write the resistance value with the correct significant figures

= (40 ± 8) 10¹ Ω