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luda_lava [24]
2 years ago
11

A mass of (200 g) of hot water at (75.0°C) is mixed with cold water of mass M at (5.0°C). The final temperature of the mixture i

s (25.0°C). What is the mass of the cold water (M)?
Physics
1 answer:
MArishka [77]2 years ago
3 0

Answer:

500 g

Explanation:

Specific heat of water = 1 j/g-c

Heat given up by hot water = 200(75-25)(1) = 10 000 j

this is the heat GAINED by the cold water

10000 j =  x ( 25 -5)(1)  

          x =500g

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kodGreya [7K]

Answer:

just calmly talk and get money to pay them the bike and explain it to them

Explanation:

4 0
3 years ago
I just need x isolated
mamaluj [8]

Answer:

x=\frac{-y+\sqrt{y^2+4rt} }{2r}

x=\frac{-y-\sqrt{y^2+4rt} }{2r}

Explanation:

rx+y=\frac{t}{x}\\\\x(rx+y)=(\frac{t}{x})x\\\\rx^2+yx=t\\\\rx^2+yx-t=t-t\\\\rx^2+yx-t=0

Solve using the quadratic formula.

x=\frac{-y+\sqrt{y^2+4rt} }{2r}

x=\frac{-y-\sqrt{y^2+4rt} }{2r}

7 0
3 years ago
Read 2 more answers
How many neutrons does element X have if its atomic number is 23 and its mass number is 90?
givi [52]

Answer:

67

Explanation:

- The atomic number (Z) of an atom is equal to the number of protons in the nucleus

- The mass number (A) of an atom is equal to the sum of protons and neutrons in the nucleus

Therefore, calling p the number of protons and n the number of neutrons, for element X we have:

Z = p = 23

A = p + n = 90

Substituting p=23 into the second equation, we find the number of neutrons:

n = 90 - p = 90 - 23 = 67

4 0
4 years ago
Read 2 more answers
Daring Darless wishes to cross the Grand Canyon of the Snake River by being shot from a cannon. She wishes to be launched at 65
guajiro [1.7K]

Answer:

She must be launched with minimum speed of <u>57.67 m/s</u> to clear the 520 m gap.

Step-by-step explanation:

Given:

The angle of projection of the projectile is, \theta =65°

Range of the projectile is, R=520 m.

Acceleration due to gravity, g=9.8\ m/s^2

The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.

The range of projectile is given as:

R=\frac{v_{0}^2\sin2\theta}{g}

Plug in all the given values and solve for minimum speed, v_0.

520=\frac{v_{0}^2\sin(2(65))}{9.8}\\520\times 9.8=v_{0}^2\sin(130)\\5096=1.532v_{0}^2\\v_0^2=\frac{5096}{1.532}\\v_0^2=3326.371\\v_0=\sqrt{3326.371}=57.67\textrm{ m/s}

Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.

3 0
3 years ago
9) A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk (M = 0.10 kg, R = 0.10 m) rotating in a horizontal
natima [27]

Answer:

a. ω₂ = 14rad/s

b. ∇K.E = 0.014J

c. The bug does not conserve force while moving on the disk (non-conservative force).

Explanation:

Mass of the bug (m) = 0.02kg

Mass of the cylindrical disk (M) = 0.10kg

Radius of the disk (r) = 0.10m

Initial angular velocity ω₁ = 10rad/s

final angular velocity ω₂ = ?

a.

To calculate the new angular velocity, we relate it to the conservation of angular momentum of the system I.e when the bug was at the edge of the disk and when it is located at the centre of the disk.

I = Mr² / 2

I₁ = Mr₂ / 2 + mr²

I₁ = moment of inertia when the bug was at the edge

I₁ = [(0.10 * 0.10²) / 2 ] + (0.02 * 0.1²)

I₁ = 0.0005 + 0.0002

I₁ = 7.0*10⁻⁴kgm²

I₂ = moment of inertia when yhe bug was at the center of the disk.

I₂ = Mr² / 2

I₂ = (0.01 * 0.01²) 2

I₂ = 0.0005kgm²

for conservation of angular momentum,

I₁ω₁ = I₂ω₂

solve for ω₂

ω₂ = (I₁ * ω₁) / I₂

ω₂ = (7.0*10⁻⁴ * 10) / 5.0*10⁻⁴

ω₂ = 14 rad/s

b. the change in kinetic energy of the system is

∇K = K₂ - K₁

∇K = ½I₂*ω₂² - ½I₁*ω₁²

∇K = ½(I₂*ω₂² - I₁ω₁²)

∇K = ½[(5.0*10⁻⁴ * 14²) - (7.0*10⁻⁴ * 10²)]

∇k = ½(0.098 - 0.07)

∇K = ½ * 0.028

∇K = 0.014J

c.

The cause of the decrease and increase in kinetic energy is because the bug uses a non-conservative force. To conserve the mechanical energy of a system, all the forces acting in it must be conservative.

The work W produced by this force brings the difference in kinetic energy of the system

W = K₂ - K₁

6 0
3 years ago
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