Where does the word 'robot' come from?

Which statement describes a step-down transformer?

natali 33 [55]

The primary coil has more turns than the secondary coil. so D.

6 0

3 years ago

a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car

Vaselesa [24]

The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 282 kg$ is the mass of the bumper car

$u_1 = 3.50 m/s$ is the initial velocity of the bumper car (we take the right as positive direction)

$v_1 = 0.800 m/s$ is the final velocity of the bumper car

$m_2 = 155 kg$ is the mass of the second bumper car

$u_2 = -1.88 m/s$ is the initial velocity of the second car (moving to the left)

$v_2$ is the final velocity of the second car

Solving for $v_2$ ,

$v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(282)(3.50)+(155)(-1.88)-(282)(0.800)}{155}=3.03 m/s$

where the positive sign means the direction is to the right.

And now we can find the momentum of the 155 kg afterwards, which is

$p_2 = m_2 v_2 = (155)(3.03)=469.7 kg m/s$ (to the right)

Learn more about momentum:

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7 0

3 years ago

A melon is projected into the air with 100 j of kinetic energy in the presence of air resistance. when it returns to its initial

MAVERICK [17]

In the presence of air resistance, a watermelon is launched into the air with 100 j of kinetic energy.

Its kinetic energy is less than 100 J when it reaches its starting point. Its kinetic energy decreases as it encounters air resistance and returns to its starting point. In actuality, some of the energy has been lost because of air resistance. Since we use the ball's original height as a point of reference, there is no potential energy when the ball is in its initial state of motion, and K is its kinetic energy. This total energy is conserved if there is no air resistance, therefore when the ball returns to its starting position, its kinetic energy will remain at 100.

Learn more kinetic energy about here:

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8 0

1 year ago

When hiking at altitude, the air pressure drops to 825,000Pa. Solve the force that site exerts against the side of a rectangular

OLga [1]

Answer:

The force is $1237500N$

Explanation:

We can apply the following expression

$pressure=\frac{Force}{Area}$

To find the area of rectangular tent wall:

$Area=length *width\\Area=2.00m*0.750m\\Area=1.5m^{2}$

Now to find the force:

$pressure=825,000pa$

$pressure=\frac{Force}{Area}$

$Force=pressure*Area\\Force=825000pa*1.5m^{2} \\Force=1237500N$

7 0

3 years ago

Calculate the energy released in joules when one mole of polonium-214 decays according to the following equation21484 Po -->

GuDViN [60]

Answer:

ΔE = 8.77 × 10¹¹ J

Explanation:

given,

²¹⁴₈₄Po -----> ²¹⁰₈₂Pb + 42 He

Atomic masses: Pb-210 = 209.98284 amu

Po-214 = 213.99519 amu

He-4 = 4.00260 amu

1 kg = 6.022 × 10²⁶ amu;

NA = 6.022 × 10²³ mol⁻¹

c = 2.99792458 × 10⁸ m/s

energy of molecule using equation

ΔE = Δm c²

Δm is mass difference and c is speed of light

Δm = 209.98284 + 4.00260 - 213.99519

Δm = - 0.00975 amu

1 amu = 1.66 x 10⁻²⁷ kg

- 0.00975 amu = - 0.00975 x 1.66 x 10⁻²⁷ Kg

= -0.016185 x 10⁻²⁷ Kg

total mass = 6.022 × 10²³ x -0.016185 x 10⁻²⁷

= -0.097467 x 10⁻⁴ Kg

ΔE = -(0.097467 x 10⁻⁴) (3 x 10^8)²

ΔE = - 8.77 × 10¹¹

ΔE = 8.77 × 10¹¹ J

8 0

4 years ago