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Leni [432]
3 years ago
11

You are heating a beaker of a chemical solution of an open flame. All of the following procedures must be followed a expect

Physics
1 answer:
pashok25 [27]3 years ago
3 0
I would say wearing rubber safety gloves.
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Which of the following correctly describes the relative air pressure at the center of a hurricane, with respect to the horizonta
vampirchik [111]

Explanation:

The pressure of hurricane which is high decreases gradually as we move higher . The pressure is maximum at the surface . Hence the relative air pressure is higher at the surface and low at the top . Like wise the pressure is low at the center and keeps on increasing when we move outside . Hence the answer is pressure is high at surface and low aloft.

6 0
3 years ago
A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle
tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

-h= vsinA\times t-gt^2/2

putting values h=15 m, v=0.8

-15 = 0.8vt - 4.9t^2  ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

-15 = 0.8\times40/0.6 - 4.9t^2

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) =  17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = -H =v sinA t - gt^2/2

⇒ 4.9t^2 - 8.9\times0.8t - 100 = 0

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0
3 years ago
Sylvia and Jadon now want to work a problem. Imagine a puck of mass 0.5 kg moving in a circular simulation. Suppose that the ten
leva [86]

Answer:

Tangential speed, v = 2.64 m/s

Explanation:

Given that,

Mass of the puck, m = 0.5 kg

Tension acting in the string, T = 3.5 N

Radius of the circular path, r = 1 m

To find,

The tangential speed of the puck.

Solution,

The centripetal force acting in the string is balanced by the tangential speed of the puck. The expression for the centripetal force is given by :

F=\dfrac{mv^2}{r}

v=\sqrt{\dfrac{Fr}{m}}

v=\sqrt{\dfrac{3.5\ N\times 1\ m}{0.5\ kg}}

v = 2.64 m/s

Therefore, the tangential speed of the puck is 2.64 m/s.

3 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
emmasim [6.3K]

2+.5 + 2.5 all over 3

5/3

1.67k/h





=2.5

4 0
3 years ago
Read 2 more answers
A baseball player throws a baseball. The ball traveled 35 yards before it landed on the ground. The ball stayed in the air for 3
Free_Kalibri [48]
Average speed = (distance traveled) / (time to cover the distance)

                         =  (35 yards)  /  (3.8 seconds)

                         =     9.21 yards per second
4 0
2 years ago
Read 2 more answers
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