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melamori03 [73]
3 years ago
11

What is 2,000m to km​

Physics
2 answers:
lesantik [10]3 years ago
7 0

Answer: 2km

Explanation:

Andru [333]3 years ago
3 0

Answer:

2km

Explanation:

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In 1896, S. Riva-Rocci developed the prototype of the current sphygmomanometer, a device used to measure blood pressure. When it
Anastasy [175]

Answer:

Explanation:

A pressure that causes the Hg column to rise 1 millimeter is called a torr. The term 1 mmHg used can replaced by the torr.

1 atm = 760 torr = 14.7 psi.

A.

120 mmHg

Psi:

760 mmHg = 14.7 psi

120 mmHg = 14.7/760 * 120

= 2.32 psi

Pa:

1mmHg = 133.322 Pa

120 mmHg = 120 * 133.322

= 15998.4 Pa

B.

80 mmHg

Psi:

760 mmHg = 14.7 psi

80 mmHg = 14.7/760 * 80

= 1.55 psi

Pa:

1mmHg = 133.322 Pa

80 mmHg = 80 * 133.322

= 10665.6 Pa

4 0
3 years ago
Until a train is a safe distance from the station it must travel at 5 m/s. Once the train is on open track it can speed up
kirza4 [7]

Answer:

I believe the answer is b

Explanation:

5 0
3 years ago
A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
4 years ago
Which TWO statements about the substances in the experiment are true?
nataly862011 [7]

Answer:

A and B are correct both are correct

4 0
3 years ago
Read 2 more answers
A third baseman makes a throw to first base 40.5 m away. The ball leaves his hand with a speed of 30.0 m/s at a height of 1.4 m
Ugo [173]

Answer:

When the ball goes to first base it will be 4.23 m high.

Explanation:

Horizontal velocity = 30 cos17.3 = 28.64 m/s

   Horizontal displacement = 40.5 m

   Time  

         t=\frac{40.5}{28.64}=1.41s          

   Time to reach the goal posts 40.5 m away = 1.41 seconds

Vertical velocity = 30 sin17.3 = 8.92 m/s

    Time to reach the goal posts 40.5 m away = 1.41 seconds

    Acceleration = -9.81m/s²

    Substituting in s = ut + 0.5at²

             s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m

    Height of throw = 1.4 m

    Height traveled by ball = 2.83 m

    Total height = 2.83 + 1.4 = 4.23 m

    When the ball goes to first base it will be 4.23 m high.

8 0
3 years ago
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