Which component of a galaxy is often found between the stars and looks like a cloud or smoke?

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

8 0

4 years ago

When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____

valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

$D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''$

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

$D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26$

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

3 0

4 years ago

As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis

klio [65]

Hi there!

We can begin by finding the acceleration of the block.

Use the kinematic equation:

$d = v_0t + \frac{1}{2}at^2$

The block starts from rest, so:

$d = \frac{1}{2}at^2\\\\12 = \frac{1}{2}a(4^2)\\\\\frac{24}{16} = a = 1.5 m/s^2$

Now, we can do a summation of forces of the block using Newton's Second Law:

$F = ma = m_bg - T$

mb = mass of the block

T = tension of string

Solve for tension:

$T = m_bg - ma = 8.2(9.8) - 8.2(1.5) = 68.06 N$

Now, we can do a summation of torques for the wheel:

$\Sigma \tau = rF\\\\\Sigma\tau = rT$

Rewrite:

$I\alpha = rT$

We solved that the linear acceleration is 1.5 m/s², so we can solve for the angular acceleration using the following:

$\alpha = a/r\\\\\alpha = 1.5/.42= 3.57 rad/sec^2$

Now, plug in the values into the equation:

$I(3.57) = (0.42)(68.06)\\\\I = (0.42)(68.06)/(3.57) = \boxed{8.00 kgm^2}$

8 0

2 years ago

Find the amount of force required to move an object of 1200 kg at a velocity of 54 km/hr?

Mkey [24]

Answer:

0 Newtons

Explanation:

The velocity of the object does not change, it is a constant 54 km/hr. When velocity does not change, acceleration is zero. Using the formula Force = mass x acceleration, we find:

mass = 1200 kg

acceleration = 0

F = (1200)(0) = 0

4 0

3 years ago

a trcuk weighs four times as much as a stationary car. if teh truck coasts into the car at 12 km/s and they stick toegther, what

Andrews [41]

Answer:

v=9.6 km/s

Explanation:

Given that

The mass of the car = m

The mass of the truck = 4 m

The velocity of the truck ,u= 12 km/s

The final velocity when they stick = v

If there is no any external force on the system then the total linear momentum of the system will be conserve.

Pi = Pf

m x 0 + 4 m x 12 = (m + 4 m) x v

0 + 48 m = 5 m v

5 v = 48

$v=\dfrac{48}{5}\ km/s$

v=9.6 km/s

Therefore the final velocity will be 9.6 km/s.

3 0

3 years ago