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marusya05 [52]
2 years ago
8

A space vehicle deploys its re–entry parachute when it's traveling at a vertical velocity of –150 meters/second (negative becaus

e the parachute is going down.) It comes to rest at 0 meters/second after 30 seconds. What's the average acceleration of the shuttle during this time span?
Physics
1 answer:
dexar [7]2 years ago
6 0

Answer:

a=5m/s^2

Explanation:

Aceleration is a change on the velocity of the object in a given time.

For this case: the initial velocity is

v_{1}=150m/s

and the final velocity is :

v_{2}=0 m/s

so, the change in velocity is:

\Delta v =v_{2}-v_{1}=0m/s - (-150m/s) =  150 m/s

and the change in time , according to the problem:

\Delta t=30s

So, the aceleration is:

a=\frac{\Delta v}{\Delta t} = \frac{150m/s}{30s} = 5m/s^2

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Hoosier Manufacturing operates a production shop that is designed to have the lowest unit production cost at an output rate of 1
abruzzese [7]

Answer:

90.77%

its capacity utilization rate for the month is 90.77%

Explanation:

The capacity utilisation rate can be expressed mathematically as;

Capacity utilisation rate = capacity used/Best operating level × 100%

Given;

Total Number of production time = 205hours

Production output/capacity used = 21400 units

Best operation rate = 115units/hour

Best operation output for the month of July( at best operation level )

=115units/hour × 205 hours = 23575 units

Capacity utilisation rate = 21400/23575 × 100%

= 90.77%

3 0
2 years ago
To a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what
neonofarm [45]
T o a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what is the man's velocity? it is 4m/s east
5 0
2 years ago
Read 2 more answers
In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being
german

Answer: Got It!

<em>Explanation:</em> Guide A Starts From Rest With Pin P At The Lowest Point In The Circular Slot, And Accelerates Upward At A Constant Rate Until It Reaches A Speed Of 175 Mm/s At The ... In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being elevated by its lead screw.

6 0
2 years ago
Calculate the number of moles in each of the following masses: 0.039 g of palladium 0.0073 kg of tantalum
marysya [2.9K]

Answer:

<em>The number of moles of palladium and tantalum are 0.00037 mole and 0.0000404 mole respectively</em>

Explanation:

Number of mole = reacting mass/molar mass

n = R.m/m.m......................... Equation 1

Where n = number of moles, R.m = reacting mass, m.m = molar mass.

For palladium,

R.m = 0.039 g and m.m = 106.42 g/mol

Substituting theses values into equation 1

n = 0.039/106.42

n = 0.00037 mole

For tantalum,

R.m = 0.0073 and m.m = 180.9 g/mol

Substituting these values into equation 1

n = 0.0073/180.9

n = 0.0000404 mole

<em>Therefore the number of moles of palladium and tantalum are 0.00037 mole and 0.0000404 mole respectively</em>

3 0
2 years ago
Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i
Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

V = 4\frac{kQ}{r}

here

r = distance of corner of the square from it center

r = \frac{a}{\sqrt2}

r = \frac{10nm}{\sqrt2} = 7.07 nm

Q = e = -1.6 * 10^{-19} C

now the net potential is given as

V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}

V = 0.815 V

now potential energy of alpha particle at this position

U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J

Now at the mid point of one of the side

Electrostatic potential is given as

V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}

here we know that

r_1 = \frac{a}{2} = 5 nm

r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}

r_2 = 11.2 nm

now potential is given as

V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}

V = -0.576 - 0.257 = -0.833 V

now final potential energy is given as

U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J

Now work done in this process is given as

W = U_f - U_i

W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}

W = -7 * 10^{-22} J

8 0
2 years ago
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