I need help on this

a student calculates experimentally the value of density of an iron as 4.4 gcm³. if the actual density of an iron is 7.6 gcm³, c

ser-zykov [4K]

Hola!

Percentage Error is a measurement of the discrepancy between an observed and a true, or accepted value.

[ refer the attachment. ]

According to Question,

% error = $\frac{7.4 - 7.6}{7.6}$ × 100

= 2.631 % = 2.7 % (approximately.)

hope it helps!

5 0

2 years ago

a ball is thrown inclined in to the air with a initial velocity of u. if it reaches the maximum height in 6 seconds , find the r

olga55 [171]

Answer:

11:1

Explanation:

At constant acceleration, an object's position is:

y = y₀ + v₀ t + ½ at²

Given y₀ = 0, v₀ = u, and a = -g:

y = u t − ½g t²

After 6 seconds, the ball reaches the maximum height (v = 0).

v = at + v₀

0 = (-g)(6) + u

u = 6g

Substituting:

y = 6g t − ½g t²

The displacement between t=0 and t=1 is:

Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]

Δy = 6g − ½g

Δy = 5½g

The displacement between t=6 and t=7 is:

Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]

Δy = (42g − 24½g) − (36g − 18g)

Δy = 17½g − 18g

Δy = -½g

So the ratio of the distances traveled is:

(5½g) / (½g)

11 / 1

The ratio is 11:1.

8 0

3 years ago

A transmission diffraction grating with 420 lines/mm is used to study the light intensity of di event orders (n). A screen is lo

Goshia [24]

Answer:

Explanation:

Diffraction grating is used to form interference pattern of dark and bright band.

Distance between adjacent slits (a ) = 1 / 420 mm

= 2.38 x 10⁻³ mm

2.38 x 10⁻⁶ m

wave length of red light

= 680 x 10⁻⁹ m

For bright red band

position x on the screen

= n λD / a , n = 0,1,2,3 etc

D = distance of screen

putting n = 1 , 2 and 3 , we can get three locations of bright red band.

x₁ = λD / a

= 680 x 10⁻⁹ x 2.8 / 2.38 x 10⁻⁶

= .8 m

= 80 cm

Position of second bright band

= 2 λD / a

= 2 x 80

= 160 cm

Position of third bright band

= 3 λD / a

= 3 x 80

= 240 cm

5 0

2 years ago

A weightlifter curls a 30 kg bar, raising it each time a distance of 0.60 m. How many times must he repeat this exercise to burn

Sholpan [36]

Answe

given,

mass of the bar, m = 30 Kg

distance of rise, h = 0.60 m

Assuming the efficiency = 25 %

energy from the pizza slice = 300 C = 1260 kJ

To consume Energy from the pizza bar is to be pulled several number of time.

( energy from pizza ) x (efficiency) = n m g h

n is the number of lift

( 1260 x 10³) x (0.25) = n x 30 x 9.8 x 0.6

$n = \dfrac{315\times 10^3}{176.4}$

n = 1786 times.

Weightlifter should lift bar 1786 times to burn off the energy.

8 0

2 years ago

In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir

Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be $S_{1}$

and the distance covered by Sir Alfred be $S_{2}$

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

$S = ut + \frac{1}{2}at^{2}$

Where $S$ is the distance traveled

$u$ is the initial velocity

$a$ is the acceleration

and $t$ is the time

For Sir George,

$S = S_{1}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.21 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\$

$t^{2} = \frac{S_{1}}{0.105}$

Now, for Sir Alfred

$S = S_{2}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.26 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\$

$t^{2} = \frac{S_{2}}{0.13}$

Since, they traveled for the same time, $t$ just before collision, we can write

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

$S_{1} + S_{2} = 86$ m

Then, $S_{2} = 86 - S_{1}$

Then,

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$ becomes

$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$

$S_{1} = 38.43$ m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0

2 years ago