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Misha Larkins [42]

2 years ago

15

true or false. because the speed of an object can change from one instant to the next, dividing the distance covered by the time

of travel gives constant speed

2 answers:

Amiraneli [1.4K]2 years ago

4 0

Answer:

False.

Explanation:

Dividing the distance covered by the time would give us the average speed, not a constant speed. Constant speed exists only if the speed of an object does not change.

scoundrel [369]2 years ago

3 0

true. Because the speed of an object can change from one instant to the next, dividing the total distance covered by the time of travel gives. average speed.

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Why is it important for a muscle to be attached to a fixed origin at one end and a moving insertion at the other? Discuss how th

vlabodo [156]

Muscles function only by contracting. This makes it necessary for one end of the muscle to be fixed and the other mobile.
Take the bicep for example.
Its origin is at the shoulder and its two heads connect to the bones of the forearm, the radius and ulna.
Now, had the muscle not been fixed at one end, and contracted, it would pull both our shoulder and forearm together resulting in an ineffective movement. The desired motion is to lift the forearm (proximal and distal movement) which can only be achieved if the bicep is fixed at the shoulder and allowed to move at the forearm.

6 0

3 years ago

A single strain gage has a nominal resistance of 120 ohm. For a quarter bridge with 120 ohm fixed resistors, strain gauge factor

natita [175]

Answer:

Output voltage is 1.507 mV

Solution:

As per the question:

Nominal resistance, R = $120\Omega$

Fixed resistance, R = $120\Omega$

Gauge Factor, G.F = 2.01

Supply Voltage, $V_{s} = 3\ V$

Strain, $\epsilon = 1000\times 10^{-6}\ strain$

Now,

To calculate the output voltage, $V_{o}$ :

WE know that strain is given by:

$\epsilon = \frac{(R + R')^{2}V_{o}}{RR'V_{s}\times G.F}$

Thus

$V_{o} = \frac{RR'V_{s}\epsilon \times G.F}{(R + R')^{2}}$

Now, substituting the suitable values in the above eqn:

$V_{o} = \frac{120\times 120\times 3\times 1000\times 10^{-6}\times 2.01}{(120 + 120)^{2}}$

$V_{o} = 1.507\ mV$

6 0

3 years ago

The student soon loses his balance and falls backwards off the board at a velocity of 1.0 m/s. Assuming momentum is conserved in

Phantasy [73]

Answer:

v2 = 27.3m/s

Explanation:

Assuming forward as positive.

Mass = m1 = 64kg

Let v be the common velocity of the student and the skateboard.

mass of skateboard = m2 = 5.94kg

v = 1.4m/s

Since the skateboard and the student are initially moving together at the same velocity their momentum together is

(m1 + m2)v

Let the final velocity of the student be v1 and the final velocity of the skateboard be v2

v1 = – 1.0m/s (falls backwards that's why the velocity is negative since we are assuming forward as positive)

Then from conservation of momentum, momentum before is equal to momentum after.

(m1 + m2)v = m1v1 + m2v2

m2v2= (m1 + m2)v – m1v1

v2 = ( (m1 + m2)v – m1v1)/m2

v2 = ( (64 + 5.94)×1.4 – 64×(-1.0))/5.94

v2 = ( (64 + 5.94)×1.4 + 64×1.0)/5.94

v2 = 27.3m/s

5 0

3 years ago

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

3 years ago

What factors might affect how fast a balloon falls to the ground?

wolverine [178]

Answer:

air resistance, gravitational force

5 0

3 years ago