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3 years ago

9,10,and11 please I want the answer I can't found it

Physics

1 answer:

Ede4ka [16]3 years ago

7 0

Q9. Inertia is the tendency of an object to resist any change in its motion.

Q10. A moving object's momentum depends on the object's mass and velocity.

Q11. I believe it would be air resistance. The fluid friction that opposes the motion of object's through the air.

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A container has 79000cm3 of mercury. Find the mass of mercury if the density is 13.6g/cm3

ZanzabumX [31]

Explanation:

Density is mass divided by volume:

D = M / V

Given V = 79000 cm³ and D = 13.6 g/cm³:

13.6 g/cm³ = M / (79000 cm³)

M = 1,074,400 g

M = 1,074 kg

Round as needed.

6 0

3 years ago

Marvin uses a long copper wire with resistivity 1.68 x 10^-8 Ω⋅m and diameter 1.00 x 10^−3 m to create a solenoid that has a 3

Mazyrski [523]

Answer with Explanation:

We are given that

Resistivity of copper wire= $\rh0=1.68\times 10^{-8}\Omega m$

Diameter=d= $1.00\times 10^{-3} m$

Radius of copper wire= $r=\frac{d}{2}=\frac{1}{2}\times 10^{-3} m$

Radius of solenoid=r' $=3 cm=3\times 10^{-2} m$

1 m=100 cm

a.Length of wire=l=11.3 m

Area of wire=A= $\pi r^2$

Where $\pi=3.14$

A= $3.14\times (\frac{1}{2}\times 10^{-3})^2$

Resistance, R= $\rho \frac{l}{A}$

Using the formula

$R=1.68\times 10^{-8}\times\frac{11.3}{3.14\times (\frac{1}{2}\times 10^{-3})^2}$

$R=0.24\Omega$

B.Length of solenoid= $2\pi r'=2\times 3.14\times 3\times 10^{-2}=0.188$ m

Number of turns= $n_0=\frac{l}{2\pi r'}=\frac{11.3}{0.188}$

$n_0$ =60

C.Potential difference,V=3 V

Current,I= $\frac{V}{R}$

I= $\frac{3}{0.24}=12.5 A$

D.Total length =0.1 m

Number of turns per unit length,n= $\frac{60}{0.1}=600$

Magnetic field along central axis inside of the solenoid,B= $\mu_0 nI$

$B=4\pi\times 10^{-7}\times 12.5\times 600=9.42\times 10^{-3} T$

4 0

3 years ago

Read 2 more answers

I WILL MARK BRAINLIEST IF CORRECT!!!

Akimi4 [234]

Answer:

I'm fairly sure that the answer is "100 m/s"

Explanation:

Fnet=ma

a=Fnet/m

a=4 N / 0.040 kg

a=100 m/s

please check my work before you submit, i don't wanna let cha down :)

5 0

3 years ago

A rectangular coil with sides 0.130 m by 0.250 m has 513 turns of wire. It is rotated about its long axis in a uniform magnetic

Sonbull [250]

Voltage = N * Δ(BA)/Δt
BA = 0.57*0.16*0.22 = 2.0064e-2 
N = 505 
115/505 = Δ(BA)/Δt = 23/101 
When the top of the coil rotates to the bottom (1/2 half cycle) BA changes from max to min and when the bottom rotates back to the top BA changes from min to max. So Δ(BA) is twice per cycle 
So 2*101Δ(BA)=23Δt and Δt = 1/f 
202*2.0064e-2/23 =Δt = 1/f => f =5.675Hz

3 0

3 years ago

A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm.

mote1985 [20]

Answer:

a) When its length is 23 cm, the elastic potential energy of the spring is

0.18 J

b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

Explanation:

Hi there!

a) The elastic potential energy (EPE) is calculated using the following equation:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretched lenght.

Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).

First, let´s convert the spring constant units into N/m:

4 N/cm · 100 cm/m = 400 N/m

EPE = 1/2 · 400 N/m · (0.03 m)²

EPE = 0.18 J

When its length is 23 cm, the elastic potential energy of the spring is 0.18 J

b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:

EPE = 1/2 · 400 N/m · (0.06 m)²

EPE = 0.72 J

When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

7 0

4 years ago