What is the tool used for classifying organisms called? A. Dichotomy B. Dichotomous key C.Dichromic key D. Periodic table

dangina [55]

Answer:

Their research differs because they were both talking about different things, Hayes was talking about how many lakes there were, while Malaska's was doing more hands on stuff like experiments. Both are important because we need to learn how the lakes formed, but we also need to do hands on experiments.

3 0

2 years ago

A Ping-Pong ball moving East at a speed of 4 m/s collides with a stationary bowling ball. The Ping-Pong ball bounces back to the

In-s [12.5K]

Answer:

They experience the same magnitude impulse

Explanation:

We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:

$p_i = p_f\\p_p + p_b = p'_p+p'_b$

where

$p_p$ is the initial momentum of the ping-poll ball

$p_b$ is the initial momentum of the bowling ball (which is zero, since the ball is stationary)

$p'_p$ is the final momentum of the ping-poll ball

$p'_f$ is the final momentum of the bowling ball

We can re-arrange the equation as follows

$p_p - p'_p = p_b'-p_b$

or

$-\Delta p_p = \Delta p_b$

which means

$|\Delta p_p | = |\Delta p_b|$ (1)

so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.

However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

$I=\Delta p$ (2)

Therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse:

$|I_p| = |I_b|$

4 0

4 years ago

3. How are meteors and meteorites different?

Natasha2012 [34]

Answer: a meteoroid is debris in space. If part of that debris comes into Earth's atmosphere, it becomes a meteor.

Explanation:

4 0

4 years ago

A 780g, 50-cm-long metal rod is free to rotate about a frictionless axle at one end. While at rest, the rod is given a short but

Semmy [17]

Answer:

9.6 rad/s

Explanation:

$L$ = length of the metal rod = 50 cm = 0.50 m

$M$ = Mass of the long metal rod = 780 g = 0.780 kg

Moment of inertia of the rod about one end is given as

$I = \frac{ML^{2}}{3} = \frac{(0.780)(0.50)^{2}}{3} = 0.065 kgm^{2}$

$F$ = force applied by the hammer blow = 1000 N

Torque produced due to the hammer blow is given as

$\tau = \frac{FL}{2}$

$\tau = \frac{(1000)(0.50)}{2}$

$\tau = 250 Nm$

$t$ = time of blow = 2.5 ms = 0.0025 s

$w$ = Angular velocity after the blow

Using Impulse-change in angular momentum, we have

$I w = \tau t\\(0.065) w = (250) (0.0025)\\w = 9.6 rads^{-1}$

5 0

3 years ago

A 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 N · m. What is the angle (assumed to

o-na [289]

Answer:

angle = 18.40 degree

Explanation:

given data

force = 95 N

distance = 0.50 m

torque = 15 N · m

to find out

angle between the wrench handle and the direction of the applied force

solution

we will apply here torque equation that is express as

torque = distance × force × sin(θ) ...................1

put here value we will get angle that is

15 = 0.50 × 95 × sin(θ)

sin(θ) = 0.315789

θ = 18.40 degree

6 0

3 years ago

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