How many nanoseconds are in one hour? How do you write the following in scientific notation?2,560,000m

Answer:

3.06 seconds time passes before the watermelon has the same velocity

watermelon going at speed 59.9 m/s

watermelon traveling when it hits the ground at speed is 79.19 m/s

Explanation:

given data

height = 320 m

speed = 30 m/s

to find out

How much time passes before the watermelon has the same velocity and How fast is the watermelon going and How fast is the watermelon traveling

solution

we will use here equation of motion that is

v = u + at    ....................1

here v is velocity 30 m/s and u is initial speed i.e zero and a is acceleration i.e 9.8 m/s²

put the value and find time t

30 = 0 + 9.8 (t)

t = 3.06 s

so 3.06 seconds time passes before the watermelon has the same velocity

and

we know superman cover distance is = velocity × time

so distance = 30 × t

and distance formula for watermelon is

distance = ut + 0.5×a×t²    .............2

here u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and h is 30 × t

30 × t = 0 + 0.5×9.8×t²

t = 6.12 s

so  by equation 1

v = u + at

v = 0 + 9.8 ( 6.12)

v = 59.9 m/s

so watermelon going at speed 59.9 m/s

and

watermelon traveling speed formula is by equation of motion

v² - u² = 2as      ......................3

here v is speed and u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and s is distance i.e 320 m

v² - 0 = 2(9.8) 320

v = 79.19 m/s

so watermelon traveling when it hits the ground at speed is 79.19 m/s

Answer:

v = 4,244,699 m/s = (4.245 × 10⁶) m/s

Explanation:

The electric force on the proton is given by

F = qE

where q = charge on the proton = (1.602 × 10⁻¹⁹) C

E = Electric field = 720,000 N/C

F = (1.602 × 10⁻¹⁹ × 720000)

F = (1.153 × 10⁻¹³) N

But this force will accelerate the proton in this magnetic field in a form of trajectory motion.

We can obtain the acceleration using Newton's first law of motion relation

F = ma

m = mass of a proton = (1.673 × 10⁻²⁷) kg

a = (F/m)

a = (1.153 × 10⁻¹³)/(1.673 × 10⁻²⁷)

a = 68,944,411,237,298 m/s²

a = (6.894 × 10¹³) m/s²

This acceleration directs the proton from the positive plate to the negative plate, covering a distance of y = 0.006 m (the distance between the plates)

Using Equations of motion, we can obtain the time taken for the proton to move from the rest at the positive plate to the negative one.

u = initial velocity of the proton = 0 m/s

y = vertical distance covered by the proton = 0.006 m

a = acceleration of the proton in this direction = (6.894 × 10¹³) m/s²

t = time taken for the proton to complete this distance = ?

y = ut + (1/2) at²

0.006 = 0 + [(1/2)×(6.894 × 10¹³)×t²]

0.006 = (3.447 × 10¹³) t²

t² = (0.006)/(3.447 × 10¹³)

t² = 1.741 × 10⁻¹⁶

t = (1.32 × 10⁻⁸) s

Then we can then calculate the minimum speed to navigate the entire length of the plates without hitting the plates.

v = ?

x = 0.056 n

t = (1.32 × 10⁻⁸)

v = (x/t)

v = (0.056)/(1.32 × 10⁻⁸)

v = 4,244,699 m/s = (4.245 × 10⁶) m/s

Hope this Helps!!!

Answer:

v = 9.09×10⁵m/s

Explanation:

Given

d = the distance between plates = 0.6cm = 0.006

E = Electric field strength = 720000N/C

m =mass of the proton = 1.67 ×10-²⁷ kg

The

Electric potential energy of the field is converted into the the kinetic energy of the proton.

So

qV = 1/2mv²

But V = Ed

So q(Ed) = 1/2mv²

v² = 2qEd/m

v = √(2qEd/m)

v = √(2×1.6×10-¹⁹×720000×0.006/1.67×10-²⁷)

v = 9.09×10⁵m/s

Answer:

15 m/s or 1500 cm/s

Explanation:

Given that

Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s

Distance moved during the hook, d(h) = 5 cm = 0.05 m

Distance moved by the fist, d(f) = 100 cm = 1 m

Average speed of the fist during the hook, v(f) = ? cm/s = m/s

This can be solved by a very simple relation.

d(f) / d(h) = v(f) / v(h)

v(f) = [d(f) * v(h)] / d(h)

v(f) = (1 * 0.75) / 0.05

v(f) = 0.75 / 0.05

v(f) = 15 m/s

Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s

Answer:

Maximum speed ( v ) = 10.4 m/s (Approx)

Explanation:

Given:

Amplitude A = 15.0 cm = 0.15 m

Frequency f = 11.0 cycles/s (Hz)

Find:

Maximum speed ( v )

Computation:

Angular frequency = 2πf

Angular frequency = 2π(11)

Angular frequency = 69.14

Maximum speed ( v ) = WA

Maximum speed ( v ) = 69.14 x 0.15

Maximum speed ( v ) = 10.371

Maximum speed ( v ) = 10.4 m/s (Approx)

The height of the building is 60 m.

calculation of building height:

The velocity of the ball should be provided by

v = u + gt

here,

u is the initial velocity of the ball = 0

v = 0 + 9.8 x 3.5

v = 34.3 m/s

Now

When the ball hits the ground, energy is conserved;

mgh = ¹/₂mv²

gh = ¹/₂v²

h = (0.5 v²) / g

h = (0.5 x 34.3²) / (9.8)

h = 60.025 m

h = 60 m

Learn more about friction here: brainly.com/question/14455351

Answer:

The height of the building is 60 m.

Explanation:

Given;

mass of the mass of the ball, m = 3 kg

time of motion, t = 3.5 s

The velocity of the ball is given by;

v = u + gt

where;

u is the initial velocity of the ball = 0

v = 0 + 9.8 x 3.5

v = 34.3 m/s

When the ball hits the ground, energy is conserved;

mgh = ¹/₂mv²

gh = ¹/₂v²

h = (0.5 v²) / g

h = (0.5 x 34.3²) / (9.8)

h = 60.025 m

h = 60 m

Therefore, the height of the building is 60 m.

Answer:4N

Explanation:

mass=4kg

Acceleration=1m/s^2

Force=mass x acceleration

Force=4 x 1

Force=4N