A 3600 lb car is traveling at a constant speed of 32 miles per hour on a straight level

In a circus act, a uniform board (length 3.00 m, mass 25.0 kg ) is suspended from a bungie-type rope at one end, and the other e

tester [92]

Answer:

Force of Rope = 122.5 N

Force of Rope = 480.2N

Explanation:

given data

length = 3.00 m

mass = 25.0 kg

clown mass = 79.0 kg

angle = 30°

solution

we get here Force of Rope on with and without Clown that is

case (1) Without Clown

pivot would be on the concrete pillar so Force of Rope will be

Force of Rope × 3m = (25kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 122.5 N

and

case (2) With Clown

so here pivot is still on concrete pillar and clown is standing on the board middle and above the centre of mass so Force of Rope will be

Force of Rope × 3m = (25kg+73kg)×(9.8ms²)×(1.5m)

solve it and we get

Force of Rope = 480.2N

4 0

4 years ago

Example of optical object

Alik [6]

Answer:

Telescope,cameras

Hope it helped

5 0

3 years ago

Read 2 more answers

A wave travels along a stretched horizontal rope. The vertical distance from crest to trough for this wave is 13 cm and horizont

shepuryov [24]

Answer:

(A) The wavelength of this wave is $56\; \rm cm$ .

(B) The amplitude of this wave is $6.5\; \rm cm$ .

Explanation:

Refer to the diagram attached. A point on this wave is at a crest or a trough if its distance from the equilibrium position is at a maximum.

The amplitude of a wave is the maximum displacement of each point from the equilibrium position. That's the same as the vertical distance between the crest (or the trough) and the equilibrium position.

On the diagram, the distance between the two gray dashed lines is the vertical distance between a crest and a trough. According to the question, that distance is $\rm 13\; \rm cm$ for the wave in this rope.
On the other hand, the distance between either gray dashed line and the black dashed line is the distance between a crest (or a trough) and the equilibrium position. That's the amplitude of this wave.

Therefore, the amplitude of the wave is exactly $\displaystyle \frac{1}{2}$ the vertical distance between a crest and a trough. Hence, for the wave in this question,

$\begin{aligned}& \text{Amplitude}\\ &= \frac{1}{2} \times (\text{Vertical distance between crest and trough}) \\ &= \frac{1}{2} \times 13\;\rm cm = 6.5\; \rm cm\end{aligned}$ .

The wavelength of a transverse wave is the same as the minimum (horizontal) distance between two crests or two troughs. That's twice the horizontal distance between a crest and a trough in the same period.

$\begin{aligned}& \text{Wavelength}\\ &= 2 \times (\text{Horizontal distance between adjacent crest and trough}) \\ &= 2 \times 28\;\rm cm = 56\; \rm cm\end{aligned}$ .

4 0

3 years ago

The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of

IRINA_888 [86]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

$T_2 = 1.008$ % higher than $T_1$

$T_2 = 0.99$ % lower than $T_1$

Explanation:

From the question we are told that

The first string has a frequency of $f_1 = 230 Hz$

The period of the beat is $t_{beat} = 0.99s$

Generally the frequency of the beat is

$f_{beat} = \frac{1}{t_{beat}}$

Substituting values

$f_{beat} = \frac{1}{0.99}$

$= 1.01 Hz$

From the question

$f_2 - f_1 = f_{beat}$ for $f_2$ having a higher tension

So

$f_2 - 230 = 1.01$

$f_2 = 231.01Hz$

From the question

$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }$

$\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}$

Substituting values

$\frac{T_2}{T_1} = \frac{(231.01)^2}{(230)^2}$

$T_2 = 1.008$ % higher than $T_1$

For $f_2$ having a lower tension

$f_1 - f_2 = f_{beat}$

So

$230 - f_2 = 1.01$

$f_2 = 230 -1.01$

$= 228.99$

From the question

$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }$

$\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}$

Substituting values

$\frac{T_2}{T_1} = \frac{(228.99)^2}{(230)^2}$

$T_2 = 0.99$ % lower than $T_1$

5 0

4 years ago

A child in a boat throws a 5.80-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 24.6kg and the ma

Sergio [31]

Answer:

-0.912 m/s

Explanation:

When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.

$(m_c + m_b)v_b + m_pv_p = 0$

where $m_c = 24.6 kg, m_b = 39 kg, m_p = 5.8 kg$ are the mass of the child, the boat and the package, respectively. $, v_p = 10m/s, v_b$ are the velocity of the package and the boat after throwing.

$(24.6 + 39)v_b + 5.8*10 = 0$

$63.6v_b + 58 = 0$

$v_b = -58/63.6 = -0.912 m/s$

3 0

4 years ago