A ball of mass 3 kg is released at the top of a track, as shown in the image below. The top of the ramp is 1 metre higher than

Time measurements from a stopwatch are not precise. Why not ?

fgiga [73]

Human error (average human reaction time is .2 seconds)

5 0

3 years ago

A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte

vaieri [72.5K]

Answer:

Part a)

$H = 44.1 m$

Part b)

$y = 13.48 m$

Part c)

$d = 8.86 m$

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

$t = \frac{v sin\theta}{g}$

now we have

$3 = \frac{vsin\theta}{9.8}$

$v sin\theta = 29.4 m/s$

Now maximum height above ground is given as

$H = \frac{v^2sin^2\theta}{2g}$

$H = \frac{29.4^2}{2(9.8)}$

$H = 44.1 m$

Part b)

Height of the fence is given as

$y = (vsin\theta) t - \frac{1}{2}gt^2$

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

$x = v_x t$

$97.5 = v_x (5.5)$

$v_x = 17.72 m/s$

now total time of flight is given as

$T = 3 + 3 = 6 s$

so range is given as

$R = v_x T$

$R = (17.72)(6)$

$R = 106.4 m$

so the distance from the fence is given as

$d = 106.4 - 97.5$

$d = 8.86 m$

7 0

4 years ago

Suppose there is a large amount of (weakly interacting) dark matter between us and a distant galaxy. How will this affect our vi

leonid [27]

Answer:

Dark matter does not affect our view, humans can see through them.

Explanation:

They do not affect our view because we can see right through the (weakly interacting) dark matter, as they do not interact or interfere with electromagnetic force.

Dark matter are often invisible substances and are difficult to spot because they don't absorb or reflect light.

7 0

3 years ago

A student calculates the density of a copper cube to be 4.15 g/cm . If the accepted value is 8.64 g/cm the percentage error in h

Helen [10]

The percentage error in his experimental value is -51.97%.

<h3>What is percentage error?</h3>

This is the ratio of the error to the actual measurement, expressed in percentage.

To calculate the percentage error of the student, we use the formula below.

Formula:

Error(%) = (calculated value-accepted value)100/(accepted............. Equation 1

From the question,

Given:

Calculated value = 4.15 g/cm
accepted value = 8.64 g/cm

Substitute these values into equation 1

Error(%) = (4.15-8.64)100/8.64
Error(%) = -4.49(100)/8.64
Error(%) = -449/8.64
Error(%) = -51.97 %

Hence, The percentage error in his experimental value is -51.97%.

Learn more percentage error here: brainly.com/question/5493941

8 0

3 years ago

State how much energy is transferred in each of the following cases: 2 grams of steam at 100 degrees Celsius condenses to water

dusya [7]

Answer:

Explanation:

When 2 gms of steam condenses to water at 100 degree latent heat of vaporization is releases which is calculated as follows

Heat released = mass x latent heat of vaporization

= 2 x 2260 = 4520 J

When 2 gms of water at 100 degree is cooled to ice water at zero degree heat is releases which is calculated as follows

Heat released = mass x specific heat x( 100-0)

= 2 x 4.2 x 100 = 840 J

When 2 gms of water at zero degree condenses to ice at zero degree latent heat of fusion is releases which is calculated as follows

Heat released = mass x latent heat of fusion

= 2 x 334 = 668 J

When 2 grams of steam at 100 degrees Celsius turns to ice at 0 degrees Celsius heat released will be sum of all the heat released as mentioned above ie

4520 + 840 +668 = 6028 J

3 0

3 years ago