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gulaghasi [49]
2 years ago
6

You are flying your Scooty Puff Jr. spaceship at 400 m/s in a direction 40 degrees from the positive y-axis towards the negative

x-axis. You're also accelerating at a rate of 0.5 m/s2 in a direction 30 degrees from the positive x-axis towards the negative y-axis. What is your current velocity?

Physics
1 answer:
vazorg [7]2 years ago
7 0

Answer:

Explanation:

Given

velocity of spaceship =400 m/s

at angle of 40^{\circ} from Positive Y axis towards negative x axis

acceleration of Spaceship=0.5 m/s^2

at an angle of 30^{\circ} from positive x axis towards the negative x axis

\vec{v}=400\left ( -\sin(40)\hat{i}+\cos (40)\hat{j}\right )

\vec{v}=-257.11\hat{i}+306.41\hat{j}

acceleration

\vec{a}=0.5\left ( \cos (30)\hat{i}-\sin (30)\hat{j}\right )

\vec{a}=0.433\hat{i}-0.25\hat{j}

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Describe several uses of plastic, and explain why plastic is a good choice for these products
miv72 [106K]
Plastic is used in many things like bags, toys, nd certain products. Plastic is very durable nd flexible :) hope i helped a bit c;
5 0
2 years ago
A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from 20.0ºC to 80.0ºC. (a) How much heatis required? Wh
Jet001 [13]

Answer:

heat used to rise temperature pan =  30.1%

heat used to rise temperature water =  69.9%

Explanation:

Given data

mass of water = 0.250 liter = 0.250 kg

aluminum pan mass = 0.500 kg

initial temperature = 20.0ºC

final temperature =  80.0ºC

to find out

heat used to rise temperature of  pan and water

solution

we find here heat transferred to the water that is

heat transferred to the water = mass of water × specific heat of water × change in temperature    ...........1

specific heat of water is 4186 J/kgºC

so

heat transferred to the water = 0.250 × 4186 × (80-20) kJ

heat transferred to the water = 62.8 kJ

and

heat transferred to the aluminum that is

heat transferred to the aluminum = mass of aluminum × specific heat of aluminum × change in temperature    ...........2

here specific heat of aluminum is 900 J/kgºC

heat transferred to the aluminum = 0.500 × 900 × (80-20) kJ

heat transferred to the aluminum = 27 kJ

so

total heat = 62.8 + 27 = 89.8 kJ

so

heat used to rise temperature pan = 27/89.8 ×100% = 30.1%

heat used to rise temperature water = 62.8 / 89.8 ×100% = 69.9%

8 0
2 years ago
A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its sta
serg [7]

Answer:

The final velocity of the bullet is 9 m/s.

Explanation:

We have,

Mass of a bullet is, m = 0.05 kg

Mass of wooden block is, M = 5 kg

Initial speed of bullet, v = 909 m/s

The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,

mv=(m+M)V\\\\V=\dfrac{mv}{m+M}\\\\V=\dfrac{0.05\times 909}{0.050+5}\\\\V=9\ m/s

So, the final velocity of the bullet is 9 m/s.

5 0
2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

Explanation:

\omega = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J

Energy used in one acceleration

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J

Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

So the number of complete accelerations is 230

8 0
2 years ago
A non-conducting sphere of radius R = 3.0 cm carries a charge Q = 2.0 mC distributed uniformly throughout its volume. At what di
BlackZzzverrR [31]

Answer:

r =3 *\sqrt{2} = 4.24 cm

Explanation:

given data

Radius of sphere 3.0 cm

charge Q = 2.0 m C

We know that maximum electric field is given as

E_{MAX}= \frac{KQ}{r^{2}}

electric field inside the sphere can be determine by using below relation

\frac{KQ}{r^{2}}= \frac{1}{2}*\frac{KQ}{R^{2}}

r = \sqrt{2}R

r =3 *\sqrt{2} = 4.24 cm

4 0
2 years ago
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