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3 years ago

Helllllllllllllllllp plss

Physics

1 answer:

andrey2020 [161]3 years ago

5 0

Answer:

I don't know answer. ssza

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List five different institutional sources of contracts that you know

dmitriy555 [2]

The contract of citizenship, the contract of merriage, the contract of loan, the contract of power, and the contract of health care. All these are examples of a contract on institutional level

6 0

3 years ago

The coefficients of friction between the load and the flatbed trailershown are μs = 0.40 and μk = 0.30. Knowing that the speed o

SOVA2 [1]

Answer:

50.97 m

Explanation:

m = Mass of truck

$\mu_s$ = Coefficient of static friction = 0.4

v = Final velocity = 0

u = Initial velocity = 72 km/h = $\dfrac{72}{3.6}=20\ \text{m/s}$

s = Displacement

Force applied

$F=ma$

Frictional force

$f=\mu_s mg$

Now these forces act opposite to each other so are equal. This is valid for the case when the load does not slide

$ma=\mu_s mg\\\Rightarrow a=\mu_s g\\\Rightarrow a=0.4\times 9.81\\\Rightarrow a=3.924\ \text{m/s}^2$

Since the obect will be decelerating the acceleration will be $-3.924\ \text{m/s}^2$

From the kinematic equations we have

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-20^2}{2\times -3.924}\\\Rightarrow s=50.97\ \text{m}$

So, the minimum distance at which the car will stop without making the load shift is 50.97 m.

5 0

3 years ago

how long (in hours) would it take a deer to run 15 kilometers if it can maintain an average velocity of 37 km/hr?

RUDIKE [14]

Use the formula t = d/v
t = 15 km/ 37 km/h
t = 0.4054
t = 0.41 h (rounded)

6 0

3 years ago

Do non metals have luster

puteri [66]

Yes, It probably true. Because the Luster it has a nonmetal, is going to be found in periodic table.

Hope it helped you.

-Charlie

4 0

3 years ago

At the end of a race a runner decelerates from a velocity of 11.00 m/s at a rate of 0.300 m/s2. (a) how far does she travel in t

inessss [21]

given that initial velocity is

$v_i = 11 m/s$

deceleration is given as

$a = -0.3 m/s^2$

now we have to find the distance covered in 16 s

$d = v_i * t + \frac{1}{2} at^2$

$d = 11*16 - \frac{1}{2}*0.3*16^2$

$d = 137.6 m$

so it will cover 137.6 m distance

part b)

in order to find the final speed

$v_f = v_i + at$

$v_f = 11 - 0.3*16$

$v_f = 6.2 m/s$

so its speed will be 6.2 m/s

8 0

3 years ago