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3 years ago

When 108 g of water at a temperature of 22.5?

1 answer:

3 0

Î”T for the first sample is the total samples final temp, minus the first sample's initial temp (47.9-22.5), so 25.4oC. Calculating q for the first sample as 108g x 4.18 J/g C x 25.4oC = 11466.58 Joules Figuring that since the first sample gained heat, the second sample must have provided the heat, so doing the calculation for the second sample, I used q=mCÎ”T 11466.58 Joules = 65.1g x 4.18 J / g C x Î”T 11466.58/(65.1gx4.18)=Î”T Î”T=42.14oC So, since second sample lost heat, it's initial temperature was 90.04oC (47.9oC final temperature of mixture + 42.14oC Î”T of second sample).

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What are the units for the spring constant, k? A. newton meters B. newton seconds C. newtons/meter D. newtons/second E. newtons/

mr Goodwill [35]

Answer:

C. N/m (newtons/meter)

Explanation:

Since the equation for potential energy for a spring is PE = 1/2kx², after you know k and x, you will get an answer in Joules.

Please let me know if you want me to explain further!

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5 0

3 years ago

Read 2 more answers

(1) A net force of 265 N accelerates a bike and rider at 2.30 m/s. What is the mass of the

kati45 [8]

Answer: 115.2kg

Explanation:

Net force = 265 N

Acceleration of bike & rider = 2.30m/s2 (The SI unit of acceleration is m/s2)

Mass of the bike and rider together = ?

Since force is the product of the mass of an object and the acceleration by which it moves, Force = Mass x Acceleration

265N = Mass x 2.30m/s2

Mass = (265N/2.30m/s2)

Mass = 115.2 kg

Thus, the Mass of the bike and rider together is 115.2kg

6 0

3 years ago

A scientist directs monochromatic light toward a single slit in an opaque barrier. The light has a wavelength of 580 nm and the

vredina [299]

Answer:

a) 9.72 mm

b) 4.86 mm

Explanation:

wave length of light λ is 580 nm = 580 \times 10⁻⁹ m

Width of slit d = 0.215\times 10⁻³ m

Distance of screen D = 1.8 m.

Width of one fringe = $\frac{\lambda\times D}{d}$

Putting the values we get fringe width

= $\frac{580\times10^{-9}\times1.8}{.000315}$

=4.86 mm.

a) Width of central maxima = 2 times width of one fringe

= 2 times 4.86

=9.72 mm

b) width of each fringe except central fringe is same , no matter what the order is.Only brightness changes .

So width of either of the two first order bright fringe will be same and it will be

= 4.86 mm.

3 0

3 years ago

A dragster starts with zero velocity and completes a 404.5 m (0.2528 mile) run in 4.922 s. If the car had a constant acceleratio

KIM [24]

Answer:

a. a=33.34ms⁻², V=164.4m/s

Explanation:

Since the dragster started with zero velocity, de determine the acceleration using of the equations of motion.

Below are the data given

Distance, s=404.5m,

time taken,t=4.922secs

Using the equation

S=ut+1/2at²

where u is the initial velocity and u=0

Making the acceleration the subject of the formula, we arrive at

a=2s/t²

a=(2*404.5)/4.922²

a=33.34ms⁻².

To determine the velocity, we use

V=u+at

V=0+33.34ms⁻² *4.922sec

V=164.4m/s

5 0

3 years ago

Read 2 more answers

A block of mass 27.00 kg sits on a horizontal surface with, coefficient of kinetic

zhannawk [14.2K]

Answer:

The force is $F = 172 \ N$

Explanation:

From the question we are told that

The mass of the block is $m_b = 27.0 \ kg$

The coefficient of static friction is $\mu_s = 0.65$

The coefficient of kinetic friction is $\mu_k = 0.50$

The normal force acting on the block is

$N = m * g$

substituting values

$N = 27 * 9.8$

$N = 294.6 \ N$

Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as

$F_f = \mu_s * N$

substituting values

$F_f = 0.65 * 264.6$

$F_f = 172 \ N$

Now for this block to move the force require is equal to $F_f$ i.e

$F= F_f$

=> $F = 172 \ N$

5 0

3 years ago