How many miles is the moon from the earth

Newtons prism experiments showed that white sunlight is made up of

DanielleElmas [232]

All the colors of the visible spectrum

4 0

2 years ago

A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20g bullets at 965m/s. the mass of the hunter (in

weqwewe [10]

When the gun is fired horizontally :

m = mass of each bullet = 4.20 g = 0.0042 kg

v = velocity of the bullet after fire = 965 m/s

M = mass of the hunter including gun = 72.5 kg

V = velocity of hunter including gun after fire = ?

V' = velocity of the combination of bullet , gun and hunter before fire = 0 m/s

Using conservation of momentum

m v + M V = (m + M) V'

(0.0042) (965) + (72.5) V = (0.0042 + 72.5) (0)

V = - 0.056 m/s

so recoil velocity comes out to be 0.056 m/s

When the gun is fired at angle 56.0⁰ above the horizontal :

m = mass of each bullet = 4.20 g = 0.0042 kg

v = velocity of the bullet after fire = 965 Cos56 = 539.62 m/s

M = mass of the hunter including gun = 72.5 kg

V = velocity of hunter including gun after fire = ?

V' = velocity of the combination of bullet , gun and hunter before fire = 0 m/s

Using conservation of momentum

m v + M V = (m + M) V'

(0.0042) (539.62) + (72.5) V = (0.0042 + 72.5) (0)

V = - 0.031 m/s

so recoil velocity comes out to be 0.031 m/s

3 0

3 years ago

11)

Vikki [24]

Answer:

salt w nulas...........

6 0

3 years ago

What would you do to improve the precision of an experiment?

GenaCL600 [577]

Explanation:

Precision represents that how close the different measurements of the sample one take are to one another.

One can increase the precision in lab by paying attention to each and every detail.
Usage of the equipment properly and also increasing the sample size.
Ensuring that the equipment is calibrated properly. They should be clean and functioning. Using equipment which is not functioning correctly can cause results to swing wildly and also bits of the debris stuck to the equipment can influence the measurements of the mass and the volume.
Each measurement must be taken multiple times, especially if experiments in which combining of the substances in specific amounts is involved.

5 0

3 years ago

Read 2 more answers

a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance

aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b) Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

= $\frac{KQ}{r^{2} }$ for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{3.10^{2} }$

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{8^{2} }$

E = - 0.77 x 10³ V/m

8 0

2 years ago