Light travels in a straight line at a constant speed of 3.0 x 10 8 m/s for 4.1

Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la

rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93 ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

$v=v_0-g \,t$

for the object's vertical displacement we have

$y-y_0=v_0\,t - \frac{g}{2} \,t^2$

If the maximum height reached by the object is given in meters, we use the value for g in $m/s^2$ which is: $9.8\,\,m/s^2$

If the maximum height of the object is given in feet, we use the value for g in $ft/s^2$ which is : $32\,\,ft/s^2$

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

$v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}$

and now we use this to express the maximum height in the second equation we typed:

$y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g}) - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}$

Then if the max height is 1.1 meters, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s$

If the max height is 1.1 feet, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s$

5 0

3 years ago

Read 2 more answers

The iron-rich core of the Moon is thought to be about how many kilometers in diameter?

natka813 [3]

The moon is thought to have an iron-rich core whose radius
is 330 km, plus or minus an uncertainty of 20 km.

That puts its diameter in the range of 620 km to 700 km.

6 0

3 years ago

How will gravitational wave discovery help the scientists in their quest to fully understand gravity?

irina [24]

<span>The heavier the body is, the stronger its gravitational pull. Just like earth, we feel gravitational pull because we are attracted to earth and so is the moon. Also, the sun is heavier than the earth and therefore, we are attracted to the sun because of its gravitational pull. When the earth revolves around the sun, both of them releases gravitational waves. Gravitational waves are ripples of waves travelling outward from the source. The more massive the orbit of two bodies, the more it emits gravitational wave. And everything around it that is near within the wave experiences a ‘pull’ toward the orbiting bodies. The advantages we get when we can measure gravitational waves are; one, we can measure the activity between two bodies in orbit in the universe, two, scientist can estimate the merging of two bodies in the universe every 15 minutes by using LIGO and three, we can know the behavior of other bodies that we did not know exist.</span>

8 0

4 years ago

Gold has a density of 19.32 g/cm3. what is the volume of a sample of gold with a mass of 27.63 grams?

schepotkina [342]

Answer: The volume is: " 1.430 cm³ " .
_____________________________
Explanation:
________________________
Volume, " V = ? ; (unknown, we need to solve for this).
Density, "D = 19.32 g / cm³ ;
mass , "m" = 27.63 g ;
_____________________________________________
The formula for density is:
____________________________________________
D = m / V ; Divide each side of the equation by: "(1/m)" ;
to isolate " V" on one side of the equation; and to solve for "V" ;
____________________________________________
1/m)*D = (1/m) * (m / V) ;
_______________________________
to get:
_______________________________
D/m = ; ↔ 1 / V ; Take the reciprocal of EACH SIDE; to isolate "V" on each side of the equation:
_____________________________
m / D = V/1 ↔ V = m / D ;
_____________________________
V = m / D ;
_______________________________
Now, plug in our given values for mass, "m" ; and Density, D"; to solve for "Volume, V " ;
_____________________________________
V = m / D = (27.63 g) ÷ (<span>19.32 g / cm</span>³) ;

= (27.63 g) * (1 cm³ / 19.32 g) ;

= (27.63 ÷ 19.32) cm³ ;

=   1.4301242236024845 cm³ ;

→ Round to "4 significant figures" ;
= 1.430 cm³ .
____________________________________________
The volume is: " 1.430 cm³ " .
______________________________________________

4 0

3 years ago

A block of mass m=16.8 kg is sliding on a surface with initial velocity v=23.2 m/s. The block has a coefficient of kinetic frict

Firdavs [7]

Answer:

t = 23.92 s

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the block and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction, downward

FN : Normal force : perpendicular to the floor, upward

fk : Kinetic friction force: parallel to the floor and opposite to the movement

F = 86.4 N , in the direction of the motion

Calculated of the W

W= m*g

W= 16.8 kg* 9.8 m/s² = 164.64 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay ay = 0

FN - W = 0

FN = W

FN = 164.64 N

Calculated of the fk

fk = μk*FN

fk = 0.426* 164.64 N

fk = 70.13 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

F - fk = m*a

86.4 -70.13 = (16.8)*(-a)

16.26 = (16.8)*(-a)

a = -(16.26 )/ (16.8)

a = - 0.97 m/s²

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula :

vf = v₀ + a*t Formula (2)

Where:

t: time interval (m)

v₀: initial speed (m/s)

vf: final speed (m/s)

Data:

v₀ = 23.2 m/s

vf = 0

a = -0.97 m/s²

Time it takes for the block to stop

We replace data in the formula (2) to calculate the time

vf= v₀+a*t

0 = 23.2+( -0.97)*t

(0.97)*t = 23.2

t = 23.2 / (0.97)

t = 23.92 s

7 0

3 years ago