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3 years ago

How much heat is needed to raise the temperature of 200 g of lead (c = 0.11 kcal/kg ∙ °c) by 10 c°?

1 answer:

3 0

Heat = mass (m)*specific heat (C)* change in temperature (Δt)

In the current scenario,
mass = 200 g = 0.2 kg
C = 0.11 kCal/kg.°C
Δt = 10 °C

Therefore,
Heat = 0.2*0.11*10 = 0.22 kCal = 0.22*4186 J = 920.92 J

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One bright summer day a swimmer is floating with her head just above the surface of a large, clear lake where the water has inde

slavikrds [6]

Answer:

5.38035 m

Explanation:

$n_w$ = Refactive index of water = 1.43

h = Depth = 5.5 m

Critical angle is given by

$\theta_c=sin^{-1}\dfrac{1}{n_w}\\\Rightarrow \theta_c=sin^{-1}\dfrac{1}{1.43}\\\Rightarrow \theta_c=44.37^{\circ}$

d = horizontal distance from the post where she no longer see the bottom of wooden post

So,

$tan\theta_c=\dfrac{d}{h}\\\Rightarrow d=tan\theta_c\times h\\\Rightarrow d=tan44.37\times 5.5\\\Rightarrow d=5.38035\ m$

The distance d is 5.38035 m

4 0

3 years ago

Why is the mechanical advantage of a bottle opener always greater than 1?

Katen [24]

For a lever, the mechanical advantage is the ratio of the lever arms. Since the lever arm of the opener is always greater than that of the output arm, the mechanical advantage is greater than 1.

6 0

3 years ago

When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and a

ryzh [129]

Complete question:

The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?

Answer:

The recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

Explanation:

Given;

combined mass of the shotgun and arm–shoulder, m₁ = 15 kg

mass of the projectile, m₂ = 0.04 kg

speed of the projectile, u₂ = 380 m/s

let the recoil velocity of the shotgun and arm–shoulder combination = u₁

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = 0

m₁u₁ = - m₂u₂

$u_1 = -\frac{m_2u_2}{m_1} \\\\u_1 = - \frac{0.04\times 380}{15} \\\\u_1 =-1.013 \ m/s\\\\u_1 = 1.013 \ m/s \ \ \ in \ opposite \ direction$

Therefore, the recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

3 0

3 years ago

A 4500 kg car accelerates from rest to 45.0

Llana [10]

The car undergoes an acceleration a such that

(45.0 km/h)² - 0² = 2 a (90 m)

90 m = 0.09 km, so

(45.0 km/h)² - 0² = 2 a (0.09 km)

Solve for a :

a = (45.0 km/h)² / (2 (0.09 km)) = 11,250 km/h²

Ignoring friction, the net force acting on the car points in the direction of its movement (it's also pulled down by gravity, but the ground pushes back up). Newton's second law then says that the net force F is equal to the mass m times the acceleration a, so that

F = (4500 kg) (11,250 km/h²)

Recall that Newtons (N) are measured as

1 N = 1 kg • m/s²

so we should convert everything accordingly:

11,250 km/h² = (11,250 km/h²) (1000 m/km) (1/3600 h/s)² ≈ 0.868 m/s²

Then the force is

F = (4500 kg) (0.868 m/s²) = 3906.25 N ≈ 3900 N

8 0

4 years ago

In a summer storm, the wind is blowing at a velocity of 8 m/s north. Suddenly in 3 seconds, the winds velocity is 23 m/s north.

cricket20 [7]

Answer:a=v-u/t

=23-8/3

=5m/s hope you got your answer

Explanation:

8 0

3 years ago