Mr Yadav buys a radio for Rs 640 and sells it for Rs 672 Find it's profit<br><br>

The theoretical line perpindicular to the surface where a light ray hits a mirror is called the

valkas [14]

That's called the "normal" to the surface at that point.

3 0

2 years ago

Platinum has a density of 21.4 g/cm3. if thieves were to steal platinum from a bank using a small truck with a maximum payload o

slega [8]

We will convert the 1dm3 in terms of cm3 as follows:

1dm^3 = (10 cm)^3
= 1000 cm^3

The mass of platinum is equal to 900 lb.
Then we will convert the mass in terms of grams as follows:
1 lb = 453.6 g
900 = 900 x 453.6 g
= 408240 g

Then density of platinum is equal to 21.4 g/cm^3
We will calculate the volume of platinum in mass 408240 g as follows:
Volume of platinum = mass of platinum / density of platinum
= 408240 g / 21.4 g/cm^3
= 19076.6 cm^3

The total volume of platinum is 19076.6 cm^3
The volume of platinum in 1 L bar is 1000cm^3
So, to calculate the number of bars we will use the formula as follows;
Number of bars = volume of platinum available / volume of platinum required in 1 L bar
= 19076.6 cm^3 / 1000 cm^3
= 19
So, the number of bars are 19.

4 0

2 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

$\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63$ seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

$D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m$ .

Now, for the maximum height, H, applying the equation of motion as

$v^2=u^2+2as$

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, $0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H$

$\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}$

$\Rightarrow H = 3.25 m$ .

Hence, the maximum height covered is 3.25 m.

8 0

2 years ago

Why would an object float in water, but sink in rubbing alcohol?

Romashka [77]

Because water is more dense than the object but rubbing alcohol is less dense than the object

6 0

3 years ago

The GPE of a 70kg person standing on a chair 1m off the ground is how many joules?

Artemon [7]

GPE I am assuming is gravitational potential energy. I'll denote it as U for simplicity.

U = mgy
U = (70kg)(9.81m/s^2)(1m) = 686.7J

U = 686.7J

Hope this helps!

8 0

3 years ago

Mr Yadav buys a radio for Rs 640 and sells it for Rs 672 Find it's profit​

Mr Yadav buys a radio for Rs 640 and sells it for Rs 672 Find it's profit