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3 years ago

A ball is thrown straight up with a velocity of 50 m/s.(use g = -10 m/s^2) a) What is the velocity in m/s after 2 seconds?

Physics

1 answer:

Andrei [34K]3 years ago

3 0

Answer:

(a) 30 m/sec

(b) -50 m/sec

Explanation:

We have given initial velocity of ball u = 50 m/sec

Acceleration due to gravity $g=-10m/sec^2$

(a) Time t = 2 sec

Now according to first equation of v = u-gt

So v=50-10×2=30 m/sec

(b) Time t = 10 sec

Now according to first equation of motion

So final velocity v = u-gt = 50-10×10 =-50 m/sec

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the apparent weight of a body wholly immersed in water is 32N and its weight in 96N and calculate volume of the body

ArbitrLikvidat [17]

Answer:

0.0065 m³

Explanation:

Apparent weight = weight − buoyancy

32 N = 96 N − (1000 kg/m³) (9.8 m/s²) V

V = 0.0065 m³

3 0

3 years ago

An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.

Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }$

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }$

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }$

$B = \boxed{7.07 *10^{-10} T}$

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

$B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}$

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

$\boxed{B = 0 T}$

3 0

2 years ago

How do P-Waves and S-Waves form?

marissa [1.9K]

P waves are produced by all earthquakes. They are compression waves that form when rocks break due to pressure in the Earth. S waves are secondary waves that are also created during an earthquake. They travel at a slower speed than the p-waves.

S waves are the waves that come after the earthquake and P waves

6 0

3 years ago

What is the kinetic energy of a 150 gram object moving at a velocity of 100 m/s?

Alex Ar [27]

Answer:

750 J

Explanation:

lets convert mass into kg first , 150 /1000 = 0.15 kg

kinetic energy = $\frac{mv^{2} }{2}$ = $\frac{0.15*100^{2} }{2} = 750 J$

3 0

3 years ago

The tallest building in the world, according to some architectural standards, is the Taipei 101 in Taiwan, at a height of 1671 f

Andrej [43]

Answer:

35.14°C

Explanation:

The equation for linear thermal expansion is $\Delta L = \alpha L_0\Delta T$ , which means that a bar of length $L_0$ with a thermal expansion coefficient $\alpha$ under a temperature variation $\Delta T$ will experiment a length variation $\Delta L$ .

We have then $\Delta L$ = 0.481 foot, $L_0$ = 1671 feet and $\alpha$ = 0.000013 per centigrade degree (this is just the linear thermal expansion of steel that you must find in a table), which means from the equation for linear thermal expansion that we have a $\Delta T =\frac{\Delta L }{\alpha L_0}$ = 22.14°. As said before, these degrees are centigrades (Celsius or Kelvin, it does not matter since it is only a variation), and the foot units cancel on the equation, showing no further conversion was needed.

Since our temperature on a cool spring day was 13.0°C, our new temperature must be $T_f=T_0+\Delta T$ = 35.14°C

3 0

3 years ago