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2 years ago

The measure of an object's speed and direction is the object's ______.

Physics

2 answers:

V125BC [204]2 years ago

3 0

Answer:

47 more rows

Explanation:

A B

speed rate of change of postion

velocity speed + direction of a moving object

weight measure of the force of gravity on an object

scale interment to measure weight

Xelga [282]2 years ago

3 0

Answer:velocity

Explanation:

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Vector c has a magnitude 24.6 m and is in the direction of the negative y-axis. vectors a and b are at angles Î± = 41.4Â° and Î²

storchak [24]

The vector c has a magnitude of 24.6m and it is in the negative y direction. Therefore
$\vec{c} = - 24.6 \hat{j}$

The vector b is 41.4° up from the x-axis. Therefore
$\vec{b} = b[cos(41.4^{o}) \hat{i} + sin(41.4^{o}) \hat{j} ] =b(0.75\hat{i} + 0.6613 \hat{j})$

The vector a is 27.7° up from the x-axis. Therefore
$\vec{a} = a[cos(22.7^{o})\hat{i} + sin(27.7^{o})\hat{j}] = a(0.8854\hat{i} + 0.4648\hat{j})$

Because $\vec{a} +\vec{b} + \vec{c} = 0$ , the sum of the x and y components should be zero. Therefore,
For the x-component,
0.8854a + 0.75b = 0
or
a + 0.847b = 0 (1)
For the y-component,
0.4648a + 0.6613b - 24.6 = 0
or
a + 1.4228b = 52.926 (2)

Subtract (1) from (2).
0.5758b = 52.926
b = 91.917
a = -0.847b = -77.854

Answer:
The magnitude of vector a is -77.85 m
The magnitude of vector b is 91.92 m

5 0

3 years ago

A combination of two identical resistors connected in series has an equivalent resistance of 12. ohms. What is the equivalent re

SIZIF [17.4K]

Answer:

R1 + R2 = R = 12 for resistors in series - so R1 = R2 if they are identical

2 R1 = 12 and R1 = R2 = 6 ohms

1 / R = 1 / R1 + 1 / R2 for resistors in parallel

R = R1 * R2 / (R1 + R2) = 6 * 6 / (6 + 6) = 3

The equivalent resistance would be 3 ohms if connected in parallel

7 0

2 years ago

A bullet of 5 gm is fired from a pistol of 1.5 kg. If the recoil velocity of pistol is 1.5 m/s,

Andreyy89

Answer:

-450 m/s

Explanation:

Momentum is conserved.

p₀ = p

0 = (1.5 kg) (1.5 m/s) + (0.005 kg) v

v = -450 m/s

6 0

3 years ago

Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s

ozzi

Answer:

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe ..... (1)

T Cosθ = mg .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

$tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}$

$tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}$

$tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}$

$tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}$

As θ is very small, so tanθ and Sinθ is equal to θ.

$\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}$

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

7 0

2 years ago

two pulse waves a and b arrive at a point in a medium simustaneously. if the amplitude of wave a is 35 units the amplitude of wa

Shalnov [3]

It depends on the type of interference.
For constructive interference, add the amplitudes to get |35 + 41| = 76 units.
For destructive, subtract them |35 - 41| = 6 units

3 0

2 years ago