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3 years ago

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The initial vertical component of the velocity is 15.3 m/s and the angle of the water is 55 degrees. What is the magnitude of th

e vector?
The answer is 18.7 m/s but how did they get this???

1 answer:

Alex787 [66]3 years ago

3 0

The vertical component of velocity is given by

V = Vy x Sin (Theta)

15.3 m/s = Vy x Sin (55 degrees)

Vy = 15.3 m/s/Sin (55 degrees)

= 15.3 m/s/0.819

= 18.681 m/s

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Calculate the magnitude of the acceleration of Io due to the gravitational force exerted by Jupiter. [Show all work, including t

galina1969 [7]

The magnitude of the acceleration due to the gravitational force exerted by Jupiter is 25.91 m/s².

<h3>What is acceleration due to gravity?</h3>

This is the acceleration of a body under a free fall due to influence of gravity.

The magnitude of the acceleration due to the gravitational force exerted by Jupiter is calculated as follows;

$F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}$

where;

M is mass of Jupiter
R is the radius of Jupiter
G is universal constant

$g = \frac{6.67 \times 10^{-11} \times 1.8986 \times 10^{27}}{(69,911 \times 10^3)^2} \\\\g = 25.91 \ m/s^2$

Thus, the magnitude of the acceleration due to the gravitational force exerted by Jupiter is 25.91 m/s².

Learn more about acceleration due to gravity here: brainly.com/question/88039

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3 years ago

You biked to the store in 10 minutes. The store was 3 km away. What was your average speed?

sineoko [7]

Answer:

speed in km/min:

0.3km/min

speed in km/h:

18km/h

speed in m/s:

5m/s

Explanation:

the average speed is defined as:

$s=\frac{d}{t}$

where $s$ is the speed, $d$ is the distance traveled, and $t$ is the time

according to the statement:

$d=3km = 3,000m$

and

$t=10min=\frac{1}{6}h=600s$ (10 minutes are equal to a sixth of an hour which is 600 seconds)

Because the units in which the speed is required are not indicated, here are some options for the answer:

Depending on the units in which you need the speed, are the quantities you should use.

Speed in km / h (kilometers per hour):

$s=\frac{3km}{\frac{1}{6}h }\\ \\s=18km/h$

speed in m / s (meters per second):

$s=\frac{3,000m}{600s}\\ s=5m/s$

or the speed in km/min (kilometers per minute)

$s=\frac{3km}{10min}\\ s=0.3km/min$

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3 years ago

At t = 0, object A is dropped from the roof of a building. At the same instant, object B is dropped from a window 10 m below the

dsp73

Answer:

Explanation:

Given two objects are dropped simultaneously

Object A is 10 m higher than object B therefore

Distance covered by object A is given by

$y_a(t)$ is given by

$y=ut+\frac{1}{2}at^2$

where y=displacement

u=initial velocity

a=acceleration

t=time

$y_a(t)=0+0.5gt^2--1$

for object B

$y_b(t)=0+0.5gt^2--2$

Subtract 1 and 2 we get

$y_a(t)-y_b(t)=0$

i.e. they will travel equal distance in equal time and distance between them remain 10 m until object B hits the ground

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3 years ago

What is the speed in meters per second of a car that is travelling at 82km/h?

4vir4ik [10]

Divide by 3.6
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3 years ago

Read 2 more answers

A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of

Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

v(t) = -25.5 cm/s

Then, it's magnitude is 25.5cm/s

d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

a(t) = -178.46Cos(3.92t)

a(t) = -178.46 Cos(3.92×0.152)

a(t) = -178.46 Cos(0.596)

a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

5 0

3 years ago