Circle the letter of each expression that has four significant figures. A. 1.25 x 10^4 B. 12.51 C. 0.0125 D. 0.1255

2. Which of the following wavelength properties would require a stopwatch to measure?

katovenus [111]

Answer:

A. wavelength

Explanation:

The others are about sound and how high it is. That has nothing to do with time.

3 0

3 years ago

Read 2 more answers

When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point

vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

$f = \frac{v}{\lambda}$

Where,

$v = Velocity \rightarrow 20m/s$

$\lambda = Wavelength \rightarrow 35*10^{-2}m$

Therefore the frequency would be given as

$f = \frac{20}{35*10^{-2}}$

$f = 57.14Hz$

The frequency is directly proportional to the angular velocity therefore

$\omega = 2\pi f$

$\omega = 2\pi *57.14$

$\omega = 359.03rad/s$

Now the maximum speed from the simple harmonic movement is given by

$V_{max} = A\omega$

Where

A = Amplitude

Then replacing,

$V_{max} = (1*10^{-2})(359.03)$

$V_{max} = 3.59m/s$

Therefore the maximum speed of a point on the string is 3.59m/s

8 0

3 years ago

The force involved in falling of an apple from a tree is known as? (a) Magnetic force (b) electrostatic force (c) Contact force

musickatia [10]

Answer:

gravitational force

...........

6 0

2 years ago

What element x is most likely to react to form the compound xf5?

djyliett [7]

antimony. It has +3,+5,-3 so yeah. the others carbon+2,+4,-4, chlorine +1,+5,+7,-1 but -1 is the most often so it isn't Cl, calcium +2.

8 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago