According to cell theory, where do new cells come from?

Skylar travels 50 meters N and then goes 30 meters W before coming straight back south 20 meters. What distance did she travel?

lesya [120]

Answer:

$100\ \text{meters}$ .

Explanation:

Distance Skylar traveled North is $50\ \text{meters}$

Then she traveled $30\ \text{meters}$ Westward.

After which she traveled $20\ \text{meters}$ towards the South.

The total distance traveled would be the sum of the distances.

$50+30+20=100\ \text{meters}$

The distance traveled by Skylar was is $100\ \text{meters}$ .

4 0

3 years ago

A 66-kg diver jumps off a 9.7-m tower. (a) Find the diver's velocity when he hits the water. (b) The diver comes to a stop 2.0 m

Mariana [72]

Answer:

(a) 13.795 m/s.

(b) -3140.28 N.

Explanation:

(a) Using newton's equation of motion,

v² = u² + 2gs.......................... Equation 1

Where v = final velocity, u = initial velocity, s = height of the tower, g = acceleration due to gravity.

Given: s = 9.7 m, u = 0 m/s ( jump from a height), g = 9.81 m/s².

Substitute into equation 1

v² = 0² + 2×9.81×9.7

v² = 190.314

v = √(190.314)

v = 13.795 m/s.

Hence the velocity of the driver when he hits the water = 13.795 m/s.

(b)

F = ma.................... Equation 2

Where F = force exerted on the diver, m = mass of the diver, a = acceleration of the diver below the water surface.

Also using

v² = u² + 2as ............ Equation 3

Note: At the point when the diver enters the water, u = 13.795 m/s, and at the point when the diver comes to a complete stop, v = 0 m/s

Given: s = 2.0 m, u = 13.795 m/s, v = 0 m/s

Substitute into equation 3

0² = 13.795²+2(2a)

0 = 190.30203 + 4a

-4a = 190.30203

a = 190.30203/-4

a = -47.58 m/s²

Also given: m = 66 kg,

Substitute into equation 3

F = (-47.58)(66)

F = -3140.28

Note: The Force is negative because it act against the motion of the diver.

Hence the net force exerted on the diver while in the water = -3140.28 N.

7 0

3 years ago

Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 44.3 ◦ . The velo

Goshia [24]

Answer:

So airplane will be 1324.9453 m apart after 2.9 hour

Explanation:

So if we draw the vectors of a 2d graph we see that the difference in angles is = 83 - 44.3 = $83-44.3=38.7^{\circ}$

Distance traveled by first plane = 730×2.9 = 2117 m

And distance traveled by second plane = 590×2.9 = 1711 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.

Using the law of cosine, $d^2$ representing the distance between the planes, we see that:

$d^2=2117^2 + 1711^2 -2\times (2117)\times (1711)cos(38.7)=1755480.2482$

d = 1324.9453 m

4 0

3 years ago

A 57 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. The acceleration of

kari74 [83]

Answer:

Her altitude as she crosses the bar, h₂ is approximately 6.1 m

Explanation:

The given parameters of the motion of the pole vaulter are;

The mass of the pole vaulter, m = 57 kg

The speed with which the pole vaulter is running, u = 11 m/s

The speed of the pole vaulter when she crosses the bar, v = 1.1 m/s

The acceleration due to gravity, g = 9.8 m/s²

From the total mechanical energy, M.E. equation, we have;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy of the motion = m·g·h

K.E. = The kinetic energy of the motion = 1/2·m·v²

By the principle of conservation of energy, we have;

The change (loss) in kinetic energy, ΔK.E. = The change (gain) in potential energy, ΔP.E.

ΔK.E. = 1/2·m·(v² - u²)

ΔP.E. = m·g·(h₂ - h₁)

Where;

h₁ = The ground level = 0 m

h₂ = The altitude with which she crosses the bar

∴ 1/2·m·(v² - u²) = m·g·(h₂ - h₁)

(h₂ - h₁) = (v² - u²)/(2·g) = (11² - 1.1²)/(2·9.8) = 6.11173469388

h₂ = 6.11173469388 + h₁ = 6.11173469388 + 0 = 6.11173469388

h₂ = 6.11173469388

Her altitude as she crosses over the bar, h₂ ≈ 6.1 m.

3 0

3 years ago

Each of two small non-conducting spheres is charged positively, the combined charge being 40 μC. When the two spheres are 50 cm

Phantasy [73]

Answer:

1.44 x 10⁻⁶ C

Explanation:

$q_{1}$ = charge on one sphere

$q_{2}$ = charge on other sphere

$q$ = Total charge on the two spheres = 40 μC

$q_{1}+ q_{2}$ = $q$

$q_{1}+ q_{2}$ = 40 x 10⁻⁶

$q_{1}$ = (40 x 10⁻⁶) - $q_{2}$ eq-1

$r$ = distance between the two spheres = 50 cm = 0.50 m

$F$ = magnitude of force between the two spheres = 2.0 N

Magnitude of force between the two spheres is given as

$F = \frac{k q_{1} q_{2}}{r^{2}}$

$2.0 = \frac{(9\times 10^{9}) ((40\times 10^{-6}) - q_{2}) q_{2}}{0.50^{2}}$

$q_{2}$ = 1.44 x 10⁻⁶ C

7 0

3 years ago