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3 years ago

15

In the opposite figure, what is the

1 answer:

kvv77 [185]3 years ago

4 0

Answer:

9

Explanation:

since the resistance is not changing

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A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of b

lbvjy [14]

Answer:

$T = 17.26 ^oC$

Explanation:

At thermal equilibrium we have heat given by aluminium must be equal to the heat absorbed by the water

so we will have

$Q_1 = Q_2$

$m_1s_1\Delta T_1 = m_2s_2\Delta T_2$

so we will have

$32.5(900)(45.8 - T) = 105.3(4186)(T - 15.4)$

so we have

$(45.8 - T) = 15.1(T - 15.4)$

so we have

$16.1 T = 277.87$

$T = 17.26 ^oC$

3 0

3 years ago

A 2300 kg sailboat is moving west at 5.5 m/s when an eastward wind

pogonyaev

The boat is initially at equilibrium since it seems to start off at a constant speed of 5.5 m/s. If the wind applies a force of 950 N, then it is applying an acceleration <em>a</em> of

950 N = (2300 kg) <em>a</em>

<em>a</em> = (950 N) / (2300 kg)

<em>a</em> ≈ 0.413 m/s²

Take east to be positive and west to be negative, so that the boat has an initial velocity of -5.5 m/s. Then after 11.5 s, the boat will attain a velocity of

<em>v</em> = -5.5 m/s + <em>a</em> (11.5 s)

<em>v</em> = -0.75 m/s

which means the wind slows the boat down to a velocity of 0.75 m/s westward.

5 0

3 years ago

A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =

Stels [109]

Answer:

628.022466 N

8.61 m/s

Explanation:

m = Mass

$\mu$ = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N$

Magnitude of frictional force is 628.022466 N

$F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2$

$v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s$

Initial speed of the player is 8.61 m/s

4 0

3 years ago

Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11

tiny-mole [99]

Answer:

The line charge density is $1.59\times10^{-4}\ C/m$

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

$V=EA$

$V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2$

$\lambda=\dfrac{V\times2\epsilon_{0}}{r}$

Where, r = radius

V = potential difference

Put the value into the formula

$\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}$

$\lambda=1.59\times10^{-4}\ C/m$

Hence, The line charge density is $1.59\times10^{-4}\ C/m$

4 0

3 years ago

A box slides down a ramp inclined at 24◦ to the horizontal with an acceleration of 1.7 m/s 2 . The acceleration of gravity is 9.

dsp73

Answer:

<h3>0.445</h3>

Explanation:

In friction, the coefficient of friction formula is expressed as;

$\mu = \frac{F_f}{R}$

Ff is the frictional force = Wsinθ

R is the reaction = Wcosθ

Substitute inti the equation;

$\mu = \frac{Wsin \theta}{W cos\theta} \\\mu = \frac{sin \theta}{cos\theta} \\\mu = tan \theta$

Given

θ = 24°

$\mu = tan 24^0\\\mu = 0.445\\$

Hence the coefficient of kinetic friction between the box and the ramp is 0.445

3 0

3 years ago