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3 years ago

5

If an object is not accelerating, then one knows for sure that it is_____.

1 answer:

Sauron [17]3 years ago

6 0

The object is maintaining a constant velocity

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An element from group 2 on the periodic table will bond lonically with elements from group 17(7A)

Natalija [7]

Answer:

Elements in group IIA have 2 electrons in their outermost shell so they can donate these electrons of group VII A which have seven electrons in their outermost shell. So it is necessary 2 elements of group VII A, each atoms accepts one electron.

Explanation:

8 0

3 years ago

a resistor in a circuit has a current through it of 0.20 A when the potential difference across is 15V. calculate the resistance

mel-nik [20]

The answer is 0.04 this is the answer you walcome

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4 years ago

What would be the mass of a 10 kg sack of sugar if it was in orbit around Saturn?

Sphinxa [80]

I would say b7.5kg for the mass of it orbiting around the Saturn because if going around something like the you could only go up in the number not go down so I say b7.5 kg of suger

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3 years ago

A 1.0-kg cart and a 0.50-kg cart sit at different positions on a low-friction track. You push on the 1.0-kg cart with a constant

Margarita [4]

Answer:

a) $W=0.8J$

b) $d_t=0.20m$

c) $\triangle K.E=0.267J$

Explanation:

From the question we are told that:

Mass of cart 1 $M_1=1.0kg$

Mass of cart 1 $M_2=0.05kg$

Force on cart 1 $F_1=4.0N$

Push Distance of cart 1 $d_1=0.20m$

Slide Distance of cart 1 $d_1'=0.35m$

a)

Generally the equation for work-done is mathematically given by

$W=f*d\\W=4*0.20\\W=0.8J \\$

b)

The systems center of mass moved a net totally of (while being pushed)

Mass 1 =0.20m

Mass 2=0

Therefore

$d_t=d_1+d_2$

$d_t=0.20+0$

$d_t=0.20m$

c)

Since work-done is equal to K.E energy of cart 1

Therefore

$W=1/2mv^2$

$V_1=\sqrt{\frac{W}{1/2m}}$

$V_1=\sqrt{\frac{0.8}{1/2(1)}}$

$V_1=1.264$

Therefore Kinetic energy before collision is

$K.E_1=1/2mv^2$

$K.E_1=1/2*1*1.264^2$

$K.E_1=0.768$

Generally from the equation for conservation of momentum the Velocity of cart 2 is mathematically given by

$v_2=\frac{m_1V_1}{m_1+m_2}$

$v_2=\frac{1*1.264}{1+0.5}$

$V_2=0.842m/s$

Therefore the final K.E is mathematically given by

$K.E_2=(1/2)(m_1+m_2)V_2^2$

$K.E_2=1/2*(1.5)(0.842)^2$

$K.E_2=0.531J$

Generally the Change in K.E is mathematically given by

$\triangle K.E=K.E_1-K.E_2$

$\triangle K.E=0.798-0.531$

$\triangle K.E=0.267J$

Therefore the will force change the kinetic energy of the system's center of mass by

$\triangle K.E=0.267J$

4 0

3 years ago

Two long straight wires are separated by a distance of d = 0.40 m. The currents are I1 = 1.0 A to the right in the upper wire an

Leni [432]

Answer:

Mangetic field, is 1 μT acting upward due to the current in the upward wire.

Explanation:

Given;

distance of separation of the two wires, d = 0.4 m

the current in the upper wire, I₁ = 1.0 A

the current in the lower wire, I₂ = 8.0 A

The magnetic field at point P, that is a distance d/2 = 0.20 m below the lower wire, is due to the current in the upper wire;

$B = \frac{\mu_oI_1}{2\pi d/2} \\\\B = \frac{4\pi *10^{-7}*1}{2\pi *0.2} = 1*10^{-6} \ T = 1 \mu T$

The direction is upward, since the magnetic field is due to the current in the upward wire.

Therefore, the mangetic field, is 1 μT acting upward due to the current in the upward wire.

7 0

3 years ago

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