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2 years ago

What is the instantaneous velocity v of the particle at t=10.0s?

1 answer:

8 0

Instantaneous velocity is defined at a specific value of time t. It can never be observed or measure. To know a velocity, one must know its distance and velocity. It is represented by the equation velocity is equal to delta distance over delta time. The problem above lacks data.

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A cliff diver on an alien planet dives off of a 32 meter tall cliff and lands in a sea of hydrochloric acid 1.20 seconds later.

Cloud [144]

Answer:

44.4m/s^2

Explanation:

Use the formula...S = ut + 1/2at^2

where...S = 32m...u = 0m/s....t = 1.20s

32 = (0)(1.20) + 0.5(1.20^2)a

;Acceleration of free fall = 44.4m/s^2

3 0

2 years ago

1. A city bus travels 6 blocks east and 8 blocks north. Each block is 100 m long. If the bus travels this distance in 15mins, wh

loris [4]

Answer:

1) 1.56 m/s

2) 1.11 m/s

Explanation:

The bus travels 6 blocks east and 8 blocks west

Each block is 100 m long

time taken to travel through this distance = 15 min

average speed of the bus = ?

the total blocks traveled = 6 + 8 = 14 blocks

total distance traveled = 14 x 100 m = 1400 m

time taken = 15 min = 15 x 60 sec = 900 sec

average speed of the bus = distance traveled/time taken

==> 1400/900 = 1.56 m/s

2) velocity = displacement/time taken

displacement is the shortest distance between the starting position and the final position of the bus.

The displacement of this bus will be the hypotenuse of the triangle formed by the motion of the bus

the distance traveled east = 6 x 100 = 600 m

the distance traveled north = 8 x 100 = 800 m

displacement = $\sqrt{800^2 + 600^2}$ = 1000 m

The velocity of the bus = 1000/900 = 1.11 m/s

4 0

2 years ago

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

Friction Force

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$

The friction forces are computed by

$\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g$

$\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g$

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

$\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g$

Simplifying

$\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)$

Plugging in the values

$\displaystyle F_{a}=0.25(9.8)[400+2(100)]$

$\boxed{F_a=1470\ N}$

8 0

2 years ago

You decide to work at a heart rate of 150 instead of 120. What area of F.I.T.T. did you change?

Rina8888 [55]

Key concepts

Heart rate

Exercising

The heart

Cardiovascular system

Health

Introduction

As Valentine's Day approaches, we're increasingly confronted with "artistic" images of the heart. Real hearts hardly resemble to two-lobed shapes adorning cards and candy boxes this time of year. And the actual shape of the human heart is important for its function of supplying blood to the entire body. You have likely noticed that your heart beats more quickly when you exercise. But have you ever taken the time to observe how long it takes to return to its normal rate after you're done exercising? In this science activity you'll get to do some exercises to explore your own heart-rate recovery time.

Background

Your heart is continuously beating to keep blood circulating throughout your body. Its rate changes depending on your activity level; it is lower while you are asleep and at rest and higher while you exercise—to supply your muscles with enough freshly oxygenated blood to keep the functioning at a high level. Because your heart is also a muscle, exercise, in turn, helps keep it healthy. The American Heart Association recommends that a person does exercise that is vigorous enough to raise their heart rate to their target heart-rate zone—50 percent to 85 percent of their maximum heart rate, which is 220 beats per minute (bpm) minus their age for adults—for at least 30 minutes on most days, or about 150 minutes a week in total. So for a 20-year-old, the maximum heart rate would be 200 bpm, with a target heart-rate zone of 100 to 170 bpm. (For those 19 or younger, target zones can vary more than they do for adults.)

i think it will help you...if it help you ...please mark brainless

8 0

2 years ago

A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.

Aleonysh [2.5K]

Answer: $53.31\°$ East of North

Explanation:

We have the following data:

Speed of the wind from East to West: $6.68 m/s$

Speed of the bee relative to the air: $8.33 m/s$

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the Speed of the wind from East to West (in the horintal part) and the speed due North relative to the ground (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

$sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}$

$sin \theta=\frac{6.68 m/s}{8.33 m/s}$

Clearing $\theta$ :

$\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})$

$\theta=53.31\°$

6 0

2 years ago