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3 years ago

The total distance traveled divided by the time it takes to travel the distance is

Physics

1 answer:

Westkost [7]3 years ago

5 0

"The total distance traveled divided by the time it takes to travel the distance"

That's actually a pretty good definition of average speed. <em>(A)</em>

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gas has a volume of 185 ml and pressure of 310 mm hg. The desiered volume is 74.0 ml. What is the required new pressure

Mamont248 [21]

Answer:

The required new pressure is 775 mm hg.

Explanation:

We are given that gas has a volume of 185 ml and a pressure of 310 mm hg. The desired volume is 74.0 ml.

We have to find the required new pressure.

Let the required new pressure be ' $\text{P}_2$ '.

As we know that Boyle's law formula states that;

$P_1 \times V_1 = P_2 \times V_2$

where, $P_1$ = original pressure of gas in the container = 310 mm hg

$P_2$ = required new pressure

$V_1$ = volume of gas in the container = 185 ml

$V_2$ = desired new volume of the gas = 74 ml

So, $P_2 = \frac{P_1 \times V_1}{V_2}$

$P_2 = \frac{310 \times 185}{74}$

= 775 mm hg

Hence, the required new pressure is 775 mm hg.

7 0

3 years ago

A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed

Vladimir [108]

Answer:

we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.

Explanation:

A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed of the rock at 2m on the way down compare with its speed at 2m on the way up?

It decreases in speed on its way down and increases in speed on its way down.

it decreases in speed on its way up because the the vertical motion is against the earths gravitational pull on an object to the earth's center

.It increases in speed on his way down because its under the influence of gravity

from newton's equation of motion we can check by

using V^2=u^2+2as

we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.

5 0

3 years ago

A diver makes 1.0 revolutions on the way from a 9.2-m-high platform to the water. Assuming zero initial vertical velocity, find

____ [38]

Answer:

Average angular velocity ≈ 4.59 rad/s

Explanation:

Using the equation of motion,

H = ut + (1/2)t² ............................ equation 1.

Where H= height, u = initial velocity(m/s), g = acceleration due to gravity(m/s²), t = time(s) u= 0 ∴ ut =0

H =(1/2)gt².................................... equation 2.

making t² the subject of the relation in equation 2,

∴ t² = 2H/g

Where H = 9.2 m, g= 9.8 m/s

∴ t² = ( 2×9.2)/9.8

t = √(2 × 9.2/9.8) = √(18.4/9.8)

t = 1.37 s.

The average angular velocity = θ/t

Where θ = is the number of revolution that the diver makes, t = time

θ = 1 rev.

Since 1 rev = 2π (rad)

t = 1.37 s

Average angular velocity = 2π/t

π = 3.143

Average angular velocity = (2×3.143)/1.37 = 6.286/1.37

Average angular velocity ≈ 4.59 rad/s

8 0

3 years ago

A can of beans that has mass M is launched by a spring-powered device from level ground. The can is launched at an angle of α0 a

scZoUnD [109]

Hi there!

Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.

Thus, the time to its highest point:
$T_h = \frac{T}{2}$

Now, we can determine the velocity at which the can was launched at using the following equation:
$v_f = v_i + at$

In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.

Therefore:
$0 = v_y + at\\\\0 = vsin\theta - g\frac{T}{2}$

***vsinθ is the vertical component of the velocity.

Solve for 'v':
$vsin(\alpha_0) = g\frac{T}{2}\\\\v = \frac{gT}{2sin(\alpha_0)}$

Now, recall that:
$W = \Delta KE = \frac{1}{2}m(\Delta v)^2$

Plug in the expression for velocity:
$W = \frac{1}{2}M (\frac{gT}{2sin(\alpha_0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{8sin^2(\alpha _0)}}$

We can use the same process as above, where T' = 2T and Th = T.

$v = \frac{gT}{sin(\alpha _0)} }\\\\W = \frac{1}{2}M(\frac{gT}{sin(\alpha _0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{2sin^2(\alpha _0)}}$

The work done in part B is 4 times greater than the work done in part A.

$\boxed{\frac{W_B}{W_A} = \frac{4}{1} = 4.0}$

4 0

2 years ago

How does the theory of plate tectonics support the theory of seafood spreading?

Zanzabum

When two tectonic plates move in opposite directions, it causes magma to rise, creating a new seafloor.

3 0

3 years ago