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Dafna11 [192]
3 years ago
15

Kind of physical barrier

Physics
1 answer:
adoni [48]3 years ago
5 0

Doors, walls, distance.

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Which of the following statements about energy in systems is true?
Elena-2011 [213]
Option number three is correct energy can be transformed and moved and released but it can't be destroyed and doesn't disappear.
8 0
3 years ago
The y-component of a projectile’s velocity is 12.1 m/s. When the projectile once again passes by the height from which it was la
Nat2105 [25]
It's 12.1 m/s, assuming that's the launch velocity that's given.
For projectile motion, velocity's y-component is parabolic/quadratic. It's x-component is constant, so you don't need to know it. 
6 0
3 years ago
In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn
kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

Acceleration of A = 1.6 m/s²

Distance = 30 m

Acceleration of B = 2.0 m/s²

We need to calculate the time for A

Using equation of motion

s=ut+\dfrac{1}{2}at_{A}^2

Put the value in the equation

30=0+\dfrac{1}{2}\times1.6\times t_{A}^2

t_{A}=\sqrt{\dfrac{30\times2}{1.6}}

t_{A}=6.12\ sec

We need to calculate the time for B

Using equation of motion

s=ut+\dfrac{1}{2}at_{B}^2

Put the value in the equation

30=0+\dfrac{1}{2}\times2.0\times t_{B}^2

t_{B}=\sqrt{\dfrac{30\times2}{2.0}}

t_{B}=5.48\ sec

We need to calculate the time for longest nap

Using formula for difference of time

t'=t_{A}-t_{B}

t'=6.12-5.48

t'=0.64\ s

Hence, B can take 0.64 sec for the longest nap .

4 0
3 years ago
The velocity of the transverse waves produced by an earthquake is 8.9 km/s, and that of the longitudinal waves is 5.1 km/s. A se
Brrunno [24]

Answer: The distance is 723.4km

Explanation:

The velocity of the transverse waves is 8.9km/s

The velocity of the longitudinal wave is 5.1 km/s

The transverse one reaches 68 seconds before the longitudinal.

if the distance is X, we know that:

X/(9.8km/s) = T1

X/(5.1km/s) = T2

T2 = T1 + 68s

Where T1 and T2 are the time that each wave needs to reach the sesmograph.

We replace the third equation into the second and get:

X/(9.8km/s) = T1

X/(5.1km/s) = T1 + 68s

Now, we can replace T1 from the first equation into the second one:

X/(5.1km/s) = X/(9.8km/s) + 68s

Now we can solve it for X and find the distance.

X/(5.1km/s) - X/(9.8km/s) = 68s

X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s

X = 68s/0.094s/km = 723.4 km

6 0
4 years ago
Please help I literally don’t understand
Veseljchak [2.6K]

Answer:

A= 2

B=3

C=4

D=5

E=7

F=8

H=12

7 0
3 years ago
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