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3 years ago

7

____ are solid while lines stretching across one or more lanes in the same direction, indicating the proper place to come to a s

top at an intersction.

1 answer:

Natalka [10]3 years ago

8 0

Stop lines are solid white lines painted across the traffic lanes at intersections and pedestrian crosswalks, indicating the exact place to stop.

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What level of nitrates is considered clean water?

lana66690 [7]

The U.S. Environmental Protection Agency (EPA) standard for nitrate in drinking water is 10 milligrams of nitrate (measured as nitrogen) per liter of drinking water (mg/L). * Drinking water with levels of nitrate at or below 10 mg/L is considered safe for everyone.

6 0

3 years ago

C) An identical spring is pulled with a force or 75 N The elastie limit of the spring is 72N

tiny-mole [99]

Answer:

Spring cannot return to its original, since a part of its deformation is <u>plastic</u>, not <u>elastic</u>.

Explanation:

Physically speaking, stress is equal to the axial force divided by effective transversal area of spring. In addition, springs have usually a linear relationship between stress and strain in <u>elastic region</u>, since they are made of ductile materials. Axial force is directly proportional to axial stress, which is also directly proportional to axial strain.

Then, if force is greater than force associated with elastic limit of the spring, then spring cannot return to its original, since a part of its deformation is <u>plastic</u>, not <u>elastic</u>.

8 0

3 years ago

Can someone please help me with these physics problems? I just don’t even know where to start.

KIM [24]

#1

for the block of mass 5 kg normal force is given as

$F_n = mg$

$F_n = 5*9.8 = 49 N$

friction force is given as

$F_f = \mu F_n$

$F_f = 0.1*49 = 4.9 N$

Net force is given as

$F_{net} = ma$

$F_{net} = 5*2 = 10 N$

now we know that

$F_{net} = F_{app} - F_f$

$10 = F_{app} - 4.9$

$F_{app} = 14.9 N$

#2

Normal force is given as

$F_n = mg$

$F_n = 6*9.8$

$F_n = 58.8 N$

now we know that

$F_{net} = F_{app} - F_f$

$F_{net} = 0$

as object moves with constant velocity

$F_{app} = F_f = 15 N$

now for coefficient of friction we can use

$F_f = \mu F_n$

$15 = \mu * 58.8$

$\mu = 0.255$

#3

net force upwards is given as

$F = 1.2 * 10^{-4} N$

mass is given as

$m = 7 * 10^{-5} kg$

now as per newton's law we can say

$F = ma$

$1.2 * 10^{-4} = 7 * 10^{-5} * a$

$a = 1.71 m/s^2$

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

$F_{net} = F_{app} - F_f$

$ma = F_{app} - F_f$

$5*6 = 40 - F_{f}$

$F_f = 40 - 30 = 10 N$

part b)

for coefficient of friction we can use

$F_f = \mu F_n$

$10 = \mu * F_n$

here normal force is given as

$F_n = mg = 5*9.8 = 49 N$

now we have

$\mu = \frac{10}{49} = 0.204$

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

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

$20 = 0 + \frac{1}{2}a(5^2)$

$a = 1.6 m/s^2$

now net force is given as

$F_{net} = ma$

$F_{net} = 10*1.6 = 16 N$

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{0 - 25}{1.5} = -16.67 m/s^2$

now the force is gievn as

$F = ma$

$F = 5*16.67 = 83.3 N$

3 0

3 years ago

Please help me to find out my mistake

Scrat [10]

From the calculation, the speed of sound at 10 K is 63.5 m/s.

<h3>What is the speed of sound?</h3>

We know that the speed of sound is directly proportional to the temperature of the body thus we can write;

V1/V2 = √T1/T2

Then;

T1 = 0 degrees or 273 K

T2 = 10 K

V1 = 330 m/s

V2 = ?

330/T2 = √273/10

330/T2 = 5.2

330 = 5.2T2

V2 = 330/5.2

V2 = 63.5 m/s

Learn more about speed of sound:brainly.com/question/15381147

#SPJ1

7 0

2 years ago

Do cats have 9 lives?

antiseptic1488 [7]

Yes cats only live up to nine years

4 0

3 years ago

Read 2 more answers