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2 years ago

11

I need help PLEASE HEEEEEEEEELPP

1 answer:

lana66690 [7]2 years ago

6 0

Answer:

a = (v2 - v1) / t

From A to B (8 - 4) m/s / 1 s = 4 m / s^2

From A to D ( 7 - 4) m/s / 5 s = .6 m / s^2

Note these equations hold for "uniform" values

They say nothing about the acceleration at intermediate points - the equation just says that his average speed increased from 4 m/s to 7 m/s during a 5 sec period

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A baseball is batted. It's a long fly ball. 4 seconds later the ball reaches the outfield 100 meters away and returns to the hei

Vitek1552 [10]

Answer:

25 m/s

Explanation:

First we should define the variables

T=4

Dx = 100

ay=-9.8

ax=0

We can use formula 1 from the BIG 5

x=(v+v0)t/2

By plugging in our variables we can get 100=4(v+v0)/2

Which is 50=v+v0

v=v0 since horizontal acceleration always equals zero

so 2v0 = 50

v0 = 25

8 0

2 years ago

Which of the following epidermal cells are the sensory receptors for touch?

never [62]

Merkel cells are the sensory receptors for touch.

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3 years ago

To test the performance of its tires, a car

Rom4ik [11]

The coefficient of static friction is 0.222

Explanation:

In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

$\mu mg=m\frac{v^2}{r}$

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where

$\mu$ is the coefficient of static friction

m is the mass of the car

g is the acceleration of gravity

v is the speed of the car

r is the radius of the track

In this problem, we have:

r = 564 m

v = 35 m/s

$g=9.8 m/s^2$

And re-arranging the equation for $\mu$ , we can find the coefficient of static friction:

$\mu = \frac{v^2}{gr}=\frac{35^2}{(9.8)(564)}=0.222$

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3 years ago

1. (30 pts) Let x(t) = cos(πt/2) be a continuous-time signal,

posledela

Given that the function of the wave is f(x) = cos(π•t/2), we have;

a. The graph of the function is attached

b. 4 units of time

c. Even

d. 4.935 J/kg

e. 1.234 W/kg

<h3>How can the factors of the wave be found?</h3>

a. Please find attached the graph of the signal created with GeoGebra

b. The period of the signal, T = 2•π/(π/2) = <u>4</u>

c. The signal is <u>even</u>, given that it is symmetrical about the y-axis

d. The energy of the signal is given by the formula;

$\frac{1}{2} \cdot \mu^{2} \cdot \omega ^{2} \cdot \: {a}^{2} \times \lambda$

Which gives;

E = 0.5 × 1.571² × 1² × 4 = <u>4.935 J/kg</u>

e. The power of the wave is given by the formula;

E = 0.5 × 1.571² × 1² × 4 × 0.25 = <u>1.234 W/</u><u>kg</u>

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8 0

2 years ago

A 4kg object has a momentum of 12 kg*m/s, what is the objects velocity?

Bezzdna [24]

Momentum = mass x velocity
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12 ÷ 4 =v
v= 3 m/s

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2 years ago