A substance has a volume of 10.0 cm3 and a mass of 89 grams. What is its density?

If electrical current is moving through a horizontal wire toward your face, what direction is the induced magnetic field?

adoni [48]

Answer:(b)

Explanation:

Given

Electric current flowing through the horizontal wire towards the observer's face.

The direction of the magnetic field is given by the right-hand thumb rule, i.e. place the thumb in the direction of current and the wrapping of fingers will give the direction of the magnetic field

the direction of the magnetic field will be counterclockwise as observed by an observer.

4 0

3 years ago

Block A, with a mass of 4.0 kg, is moving with a speed of 2.0 m/s while block B , with a mass of 8.0 kg, is moving in the opposi

Anon25 [30]

Consider velocity to the right as positive.

First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right

Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left

Total momentum of the system is
P = m₁v₁ + m₂v₂
= 4*2 + 8*(-3)
= -16 (kg-m)/s

Let v (m/s) be the velocity of the center of mass of the 2-block system.

Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s

Answer:
The center of mass is moving at 1.33 m/s to the left.

5 0

3 years ago

Starting from rest, a small block of mass m slides frictionlessly down a circular wedge of mass M and radius R which is placed o

guapka [62]

Answer:

Part a)

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

Explanation:

PART A)

As we know that there is no external force on the system of two masses in horizontal direction

So here the two masses will have its momentum conserved in horizontal direction

So we have

$mv + MV = 0$

Also we know that here no friction force on the system so total energy will always remains conserved

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}MV^2 = mgR$

now we have

$\frac{1}{2}m(\frac{MV}{m})^2 + \frac{1}{2}MV^2 = mgR$

$\frac{1}{2}MV^2(\frac{M}{m} + 1) = mgR$

so we have

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

and another block has speed

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

7 0

3 years ago

SOMEONE PLEASE HELP ME ASAP PLEASE!!!!!

Zina [86]

Answer:

x₁ = 58.09 m

Explanation:

Displacement is calculated by finding the final distance away from a point then subtracting the initial distance.

Given that initial position of object is x=25.89 m

Displacement Δx=32.2 m

Final position x₁=?

Δx = x₁-x

32.2= x₁-25.89

32.2+25.89 = x₁

58.09 m =x₁

5 0

3 years ago

Divide 0.0081 by 300 seconds and express the answer in scientific notation.

Colt1911 [192]

Answer:

4×10⁴ seconds

Explanation:

Assuming you have a calculator on hand, inputting 300 ÷ .0081 give should give you 37037.0370...

In scientific notation, that number would be 3.70370370...×10⁴ and when taking into account the rounding rules for significant figures, the answer is 4×10⁴ seconds.

3 0

3 years ago