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2 years ago

What is the final temperature when a 3.0 kg gold bar at 99 0C is dropped into 0.22 kg of water at 25oC?

Physics

1 answer:

slavikrds [6]2 years ago

8 0

I will post my work, but is that 99 degrees Celsius and 25 degrees Celsius?

All you have to do is plug in the initial temperature for gold where it says Tg and the initial temperature for the water where it says Tw and then plug that in and you will have your answer.

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A 150 g particle at x = 0 is moving at 8.00 m/s in the +x-direction. As it moves, it experiences a force given by Fx=(0.850N)sin

krok68 [10]

Answer:

9.98 m/s

Explanation:

The force acting on the particle is defined by the equation:

$F=(0.850) sin (\frac{x}{2.00})$ [N]

where x is the position in metres.

The acceleration can be found by using Newton's second law:

$a=\frac{F}{m}$

where

m = 150 g = 0.150 kg is the mass of the particle. Substituting into the equation,

$a=\frac{0.850}{0.150}sin (\frac{x}{2.00})=5.67 sin(\frac{x}{2.00})$ [m/s^2]

When x = 3.14 m, the acceleration is:

$a=5.67 sin(\frac{3.14}{2.00})=5.67 m/s^2$

Now we can find the final speed of the particle by using the suvat equation:

$v^2-u^2=2ax$

where

u = 8.00 m/s is the initial velocity

v is the final velocity

$a=5.67 m/s^2$

x = 3.14 m is the displacement

Solving for v,

$v=\sqrt{u^2+2ax}=\sqrt{8.00^2+2(5.67)(3.14)}=9.98 m/s$

And the speed is just the magnitude of the velocity, so 9.98 m/s.

4 0

2 years ago

Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the b

Salsk061 [2.6K]

Answer:

Explanation:

GIVEN DATA:

Distance between keisha and her friend 8.3 m

angle made by keisha toside building 30 degree

height of her friend monique is 1.5 m

from the figure

$\Delta ACB$

$tan 30 = \frac{8.3}{h}$

$h= \frac{8.3}{tan 30} = 14.376 m$

therefore

height of keisha is $= h + 1.5 m$

= 14.376 + 1.5

$= 15.876 \simeq 16 m$

therefore option c is correct

5 0

2 years ago

A car increases its speed from 9.6 meters per second to 11.2 meters per second in 4.0 seconds. The average acceleration of the c

Gwar [14]

The acceleration is the change of speed/velocity over time. Thus to calculate this you do (V1-V2)/T or (11.2-9.6)/4 or 0.4 m/s^2

5 0

2 years ago

Read 2 more answers

Your employer asks you to build a 18-cmcm-long solenoid with an interior field of 5.8 mTmT . The specifications call for a singl

andrezito [222]

Answer:

Explanation:

The magnetic field in a solenoid is

B = μ₀ N / L I

Where N is the number of turns, L the solenoid length and I the current

N = B L / μ₀ I

Let's calculate

N = 5.8 10⁻³ 0.18 / 4 π 10⁻⁷ 1

N = 8.3 102 laps

N = 831 laps

Let's find the solenoid length

For this we use a rule of proportions

L_solenoid = Turns * wire diameter

L_ solenoid = 831 * 0.41 10--3

L_solenoid = 0.3407 m

We see that two turns are needed in the wire to have a length of 0.18 m

5 0

2 years ago

Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete

Margarita [4]

Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(15)^2}{22.5}$

$a=10\ m/s^2$

We know that, $g=9.8\ m/s^2$

$a=\dfrac{10\times g}{9.8}$

$a=1.02\ g$

Hence, this is the required solution.

5 0

2 years ago