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3 years ago

How much mili ampere is 2 ampere

Physics

1 answer:

Rasek [7]3 years ago

8 0

Answer:

2000 mili ampere

Explanation:

1 ampere is = to 1000 miliampere so 2 x 1000 is equal to 2000 miliampere

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A stationary 90 kg baseball pitcher throws a 0.15 kg baseball forward

Marina CMI [18]

Answer: momentum = 6kgm/s

Explanation:

given that the baseball pitcher is at stationary position, his velocity will be equal to zero. If velocity is zero, his linear momentum will therefore equal to zero.

Linear momentum is the product of mass and velocity. Given that the baseball has

Mass M = 0.15 kg

Velocity V = 40 m/s

Momentum = MV

Momentum = 0.15 × 40 = 6 kgm/s

4 0

3 years ago

A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$

The change in momentum of the astronaut is

$\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s$

And the duration of the push is

$\Delta t = 0.600 s$

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

$F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N$

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

$K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J$

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

$K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J$

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4 years ago

A speed does not involve the element of

guajiro [1.7K]

A speed does not involve the element of direction.

6 0

3 years ago

Ice core samples are used to measure…

blsea [12.9K]

Answer:

the correct answer is A

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3 years ago

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Enterprising students set an enormous slip-n-slide (a plastic sheetcovered in water to reduce friction) on flat ground. If the s

Klio2033 [76]

Answer:

a = -0.05 m/s² (negative sign shows deceleration)

Explanation:

In order, to find out the minimum average acceleration for a student starting at 5 m/s to slide to the end, we can use 3rd equation of motion. 3rd equation of motion is given as follows:

2as = Vf² - Vi²

where,

a = minimum acceleration required = ?

s = minimum distance covered = 250 m

Vf = Final Speed = 0 m/s (for minimum acceleration the student will barely cover 250 m and then stop)

Vi = Initial Velocity = 5 m/s

Therefore,

2a(250 m) = (0 m/s)² - (5 m/s)²

a = - (25 m²/s²)/(500 m)

<u>a = -0.05 m/s²</u> (negative sign shows deceleration)

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3 years ago