What kind of relationship exists between sound and the temperature of material

If the car’s speed decreases at a constant rate from 64 mi/h to 30 mi/h in 3.0 s, what is the magnitude of its acceleration, ass

mixas84 [53]

Answer: $3.874 m/s^2$

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

$60mi/h \approx 26.8224m/s$

$34mi/h \approx 15.1994 m/s$

we know acceleration is given by $=\frac{velocity}{Time}$

$a=\frac{15.1994-26.8224}{3}$

$a=-3.874 m/s^2$

negative indicates that it is stopping the car

Distance traveled

$v^2-u^2=2as$

$\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s$

$s=\frac{488.419}{2\times 3.874}$

s=63.038 m

7 0

3 years ago

In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 2.15 kg. The

Maksim231197 [3]

Answer:

0.339 kgm²

Explanation:

We know the period of this pendulum, T = 2π√(I/mgh) where I = moment of inertia of the object about the pivot axis, m = mass of object = 2.15 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.163 m.

Since T = 2π√(I/mgh), making I subject of the formula, we have

I = mghT²/4π²

Now since it takes 241 s to complete 113 cycles, then it takes 241 s/113 cycles to complete one cycle.

So, T = 241 s/113 = 2.133 s

So, Substituting the values of the variables into I, we have

I = mghT²/4π²

I = 2.15 kg × 9.8 m/s² × 0.163 m × (2.133 s)²/4π²

I = 15.63/4π² kgm²

I = 0.396 kgm²

Now from the parallel axis theorem, I = I' + mh² where I' = moment of inertia of object with respect to its center of mass about an axis parallel to the pivot axis

I' = I - mh²

I' = 0.396 kgm² - 2.15 kg × (0.163 m)²

I' = 0.396 kgm² - 0.057 kgm²

I' = 0.339 kgm²

7 0

3 years ago

A small plane flies 40.0 km in a direction 60° north of east and then flies 30.0 km in a direction 15° north of east. Use the an

lana66690 [7]

Answer:

d= 64.7 km

$\theta = 40.9^{o}$

displacement vector $=r_xi + r_yj = 48.9i + 42.4j$

Explanation:

total distance = 40 + 30 = 70 km

during 1st flight

$r_1 x = 40*cos60$

$r_1 x = 20 km$

$r_1 y = 40*sin60$

$r_1 y = 34.64 km$

during 2nd flight

$r_2 x = 30*cos15$

$r_2 x = 28.9 km$

$r_2 y = 30*sin15$

$r_2 y = 7.76 km$

the two component of r are:

$r_x = r_1x + r_2x = 20 + 28.9 = 48.9 km$

$r_y = r_1y + r_2y = 34.64 + 7.76 = 42.4 km$

Geographical Direction $\theta = tan^{-1}\frac{r_y}{r_x} [tex]\theta = 40.9^{o}$

Displacement d $= \sqrt{r_x^2 + r_y^2}$

$d = \sqrt{48.9^2+42.4^2} = 64.7 km$

d= 64.7 km

displacement vector $=r_xi + r_yj = 48.9i + 42.4j$

8 0

3 years ago

The earth's radius is about 4000 miles. Kampala, the capital of Uganda, and Singapore are both nearly on the equator. The distan

Cerrena [4.2K]

Answer:

a) the required angle in both radian and degree is 1.25 rad and 71.6°

b) the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec

Explanation:

Given the data in the question;

a)

we know that The expression for the angle subtended by an arc of circle at the center of the circle is,

θ = Length / radius

given that Length is 5000 miles and radius is 4000 miles

we substitute

θ = 5000 miles / 4000 miles

θ = 1.25 rad

Radian to Degree

θ = 1.25 rad × ( 180° / π rad )

θ = 71.6°

Therefore, required angle in both radian and degree is 1.25 rad and 71.6°

b)

The flight from Kampala to Singapore take 9 hours.

the plane's angular speed relative to the earth = ?

we know that, the relation between angular velocity and angular displacement is;

ω = θ / t

given that θ is 1.25 rads and time t is 9 hours or ( 9 × 3600 sec ) = 32400 sec

we substitute

ω = 1.25 rad / 32400 sec

ω = 3.86 × 10⁻⁵ rad/sec

Therefore, the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec

5 0

3 years ago

At the same moment, one rock is thrown upward at 4.5 m/s and another thrown downward at 6.2 m/s. What is the relative velocity o

erastova [34]

The correct answer is
<span>C) -10.7 m/s

In fact, the first rock is moving upward with velocity +4.5 m/s, while the second rock is moving downward with velocity -6.2 m/s, with respect to a fixed reference frame. In the reference frame of the first rock, instead, the second rock is moving with velocity equal to its velocity in the fixed frame minus the velocity of the reference frame of the first rock:
</span> $v=-6.2 m/s -(+4.5 m/s) = -10.7 m/s$ <span>
</span>

8 0

3 years ago