What is one defining feature of a prokaryotic cell

Two pebbles, Pebble A and Pebble B, are thrown horizontally with the same force. Pebble A's mass is 3 times the

Hitman42 [59]

Answer:

Pebble A has 1/3 the acceleration as pebble B.

Explanation:

F = m×a

mass of a = 3 × mass of b (m_a = 3 × m_b)

Same starting force, F

m_a = mass of a

m_b = mass of b

a_a = acceleration of a

a_b = acceleration of b

F = m_a × a_a = m_b × a_b

3 × m_b × a_a = m_b × a_b

3 × a_a = a_b

OR

a_a = a_b / 3

5 0

3 years ago

__________energy might also be released during a chemical reaction

pentagon [3]

Kinetic energy i think

7 0

3 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

3 years ago

A boat travels at 30 mph for a huge, solid cliff that is about 3,000 meters away. When the horn on the boat makes a toot, you ca

AleksandrR [38]

Answer:The frequency of the echo is slightly decreased

Explanation:

Given

speed of boat $=30\ mph$

cliff is $3000\ m$ away

when boat is still , suppose t is the time taken by the echo to reach observer on the boat

But as soon as boat starts moving the distance between cliff and boat decreasing and time for echo to reach observer also decreases

and we know $time \propto \frac{1}{frequency}$

therefore frequency of the echo slightly decreased.

5 0

3 years ago

A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.

LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

$\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

$W = \frac{1}{2}(50)(5.61^2) = 786.8025 J$

Now, we can define work:

$\large\boxed{W = Fdcos\theta}}$

Now, plug in the values:

$786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta$

Solve for theta:

$cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}$

4 0

3 years ago