What is the displacement of an object during a specific unit of time?

Janice has just measured the density of an object. Which value is possible? (Density: D = )

babunello [35]

It is 6 g/cm3 because density cannot be negative, and it is not speed in which the unit would be m/s.

6 0

3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

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

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago

If five joules were required to move a crate in 3.7 seconds, what power was applied?

AleksAgata [21]

Answer:

The answer to your question is 1.35 Watts

Explanation:

Data

Work = W = 5 J

time = t = 3.7 s

Power = P = ?

Formula

Power is a rate in which work is done or energy is transferred over time

P = $\frac{W}{t}$

Substitution

$P = \frac{5}{3.7}$

Result

P = 1.35 W

8 0

3 years ago

A student evaluates a weight loss program by calculating the number of times he would need to climb a 14.0 m high flight of step

Liula [17]

Answer:

400 trips

Explanation:

Mechanical energy needed to climb 14 m by a man of 68 kg

= mgh

= 68 x 9.8 x 14

= 9330 J

1 Kg of fat releases 3.77 x 10⁷ J of energy

.45 kg of fat releases 1.6965 x 10⁷ J of energy

22% is converted into mechanical energy

so 22% of 1.6965 x 10⁷ J

= 3732.3 x 10³ J of mechanical energy will be available for mechanical work.

one trip of climbing of 14 m requires 9330 J of mechanical energy

no of such trip possible with given mechanical energy

= 3732.3 x 10³ / 9330

= 400 trips

7 0

3 years ago

Joe and Max shake hands and say goodbye. Joe walks east 0.50 km to a coffee shop, and Max flags a cab and rides north 3.45 km to

timama [110]

Answer:

3.486 km

Explanation:

Suppose Joe and Max's directions are perfectly perpendicular (east vs north). We can calculate their distance at the destinations using Pythagorean theorem:

$s = \sqrt{J^2 + M^2}$

where J = 0.5 km and M= 3.45 km are the distances between Joe and Max to their original parting point, respectively. s is the distance between them.

$s = \sqrt{0.5^2 + 3.45^2} = \sqrt{12.1525} = 3.486 km$

8 0

3 years ago