Một vật được ném lên trên theo phương thẳng đứng. Người quan sát

When resting, a person has a metabolic rate of about 3.0 105 joules per hour. The person is submerged neck-deep into a tub conta

jek_recluse [69]

Answer:

The temperature after half an hour is $19.3002^{\circ}$

Solution:

As per the question;

Metabolic rate of the person is 3.0105 J/h

Temperature, T = $19.30^{\circ}$

Mass of the water, $m_{w} = 1.2103 kg$

Time duration, t = 0.5 h = 30 min = 180 s

Now,

Heat, $Q = ms\Delta t$

Thus heat transfer in half an hour:

$Q = 3.0105\times 0.5 = 1.505 J$

Now, the temperature of water after half an hour, T' is given by:

$Q = m_{w}s\Delta T = ms(T' - T)$

where

s = 4186 J

$1.505 = 1.21103\times 4186\times (T' - 19.30^{\circ})$

$T' = 19.3002 ^{\circ}$

4 0

3 years ago

What are the relationships between position, velocity, and acceleration as it relates to time?

Ber [7]

Velocity = distance /time
acceleration = velocity / time

5 0

3 years ago

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An accelerating voltage of 2.47 x 10^3 V is applied to an electron gun, producing a beam of electrons originally traveling horiz

Dmitry [639]

Answer:

6.3445×10⁻¹⁶ m

Explanation:

E = Accelerating voltage = 2.47×10³ V

m = Mass of electron

Distance electron travels = 33.5 cm = 0.335 cm

$E=\frac{mv^2}{2}\\\Rightarrow v=\sqrt{\frac{2E}{m}}\\\Rightarrow v=\sqrt{\frac{2\times 2470\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=29455356.08671\ m/s$

Deflection by Earth's Gravity

$\Delta =\frac {gt^2}{2}$

Now, Time = Distance/Velocity

$\Delta =\frac {g\frac{s^2}{v^2}}{2}\\\Rightarrow \Delta =\frac{9.81\frac{0.335^2}{29455356.08671^2}}{2}\\\Rightarrow \Delta =6.3445\times 10^{-16}\ m$

∴ Magnitude of the deflection on the screen caused by the Earth's gravitational field is 6.3445×10⁻¹⁶ m

3 0

3 years ago

You drag a suitcase of mass 8.2 kg with a force of f at an angle 41.9 ◦ with respect to the horizontal along a surface with kine

DedPeter [7]

Answer:

35.6 N

Explanation:

We can consider only the forces acting along the horizontal direction to solve the problem.

There are two forces acting along the horizontal direction:

- The horizontal component of the pushing force, which is given by

$F_x = F cos \theta$

with $\theta=41.9^{\circ}$

- The frictional force, whose magnitude is

$F_f = \mu mg$

where $\mu=0.33$ , m=8.2 kg and g=9.8 m/s^2.

The two forces have opposite directions (because the frictional force is always opposite to the motion), and their resultant must be zero, because the suitcase is moving with constant velocity (which means acceleration equals zero, so according to Newton's second law: F=ma, the net force is zero). So we can write:

$F_x - F_f=0\\F_x = F_f\\F cos \theta = \mu mg\\F=\frac{\mu mg}{cos \theta}=\frac{(0.33)(8.2 kg)(9.8 m/s^2)}{cos(41.9^{\circ})}=35.6 N$