Planet with the Great Red Spot

Which particle is most likely to interact with your hand? Select one: a. Alpha particle b. Beta particle c. Gamma particle d. Ne

IgorLugansk [536]

Answer:

The correct option is a

Explanation:

The alpha particle has the lowest penetrating power of the trio of alpha, beta and gamma particles and can be stopped by a sheet of paper and hence cannot penetrate a human skin. Beta particle has a higher penetrating power than alpha particle (some of it penetrates the human skin and some do not) while the gamma particle has the highest penetrating power (with all of it penetrating the human skin).

From the above description, it can be deduced that the alpha particle will stay and interact with the hand (because of its low penetrating power) as the remaining particles move through the skin.

4 0

2 years ago

A system gains 1500 J of heat, while the internal energy of the system increases by 4500 J and the volume decreases by . Assume

Assoli18 [71]

Answer:

Hence the pressure is $3\times 10^5 Pa$

Explanation:

Given data

Q=1500 J system gains heat

ΔV=- 0.010 m^3 there is a decrease in volume

ΔU= 4500 J internal energy decrease

We know work done is

W= Q- ΔU

=1500-4500= -3000 J

The change in the volume at constant pressure is

ΔV= W/P

there fore P = W/ΔV= -3000/-0.01= 3×10^5

Hence the pressure is $3\times 10^5 Pa$

3 0

2 years ago

At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers do not f

worty [1.4K]

Answer:

$v_{min} \approx 17.153\,\frac{m}{s}$

Explanation:

The roller coaster begins with maximum kinetic energy and no gravitational potential energy. The gravitational potential energy reaches its maximum when roller coaster is upside down at the top of the circle. The physical model for the roller coaster is constructed by means of the Principle of Energy Conservation:

$\frac{1}{2}\cdot m \cdot v_{min}^{2} = m\cdot g \cdot h$

The minimum velocity is:

$v_{min} = \sqrt{2\cdot g \cdot h}$

Let assume that radio of curvature is measured in meters. Hence:

$v_{min} = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot(15\,m)}$

$v_{min} \approx 17.153\,\frac{m}{s}$

8 0

2 years ago

Hy would you expect sodium (Na) to react strongly with chlorine (Cl)?

erastova [34]

Chlorine needs to gain one electron and sodium needs to lose one electron

8 0

2 years ago

An 18.8 kg block is dragged over a rough, horizontal surface by a constant force of 156 N acting at an angle of angle 31.9◦ abov

Sav [38]

Answer:

a) W = 2635.56 J

b) Wf = 423.27 J

c) c) The Sign of the work done by the frictional force (Wf) is negative (-)

d) W=0

Explanation:

Work (W) is defined as the Scalar product of the force (F) by the distance (d) that the body travels due to this force .

The formula for calculate the work is :

W = F*d*cosα

Where:

W : work in Joules (J)

F : force in Newtons (N)

d: displacement in meters (m)

α :angle that form the force (F) and displacement (d)

Known data

m = 18.8 kg : mass of the block

F= 156 N,acting at an angle θ = 31.9◦°: angle above the horizontal

μk= 0.209 : coefficient of kinetic friction between the cart and the surface

g = 9.8 m/s²: acceleration due to gravity

d = 19.9 m : displacement of the block

Forces acting on the block

We define the x-axis in the direction parallel to the movement of the cart on the floor and the y-axis in the direction perpendicular to it.

W: Weight of the cart : In vertical direction downaward

N : Normal force : In vertical direction the upaward

F : Force applied to the block

f : Friction force: In horizontal direction

Calculated of the weight of the block

W= m*g = ( 18.8 kg)*(9.8 m/s²)= 184.24 N

x-y components of the force F

Fx = Fcosθ = 156 N*cos(31.9)° = 132.44 N

Fy = Fsinθ = 156 N*sin(31.9)° = 82.44 n

Calculated of the Normal force

Newton's second law for the block in y direction :

∑Fy = m*ay ay = 0

N-W+Fy= 0

N-184.24+82,44= 0

N = 184.24-82,44

N = 101.8 N

Calculated of the kinetic friction force (fk):

fk = μk*N = (0.209)*( 101.8)

fk = 21.27 N

a) Work done by the F=156N.

W = (Fx) *d *cosα

W = (132.44 )*(19.9)(cos0°) (N*m)

W = 2635.56 J

b) Work done by the force of friction

Wf = (fk) *d *cos(180°)

Wf = (21.27 )*(19.9) (-1) (N*m)

Wf = - 423.27 J

Wf = 423.27 J :magnitude

c) The Sign of the work done by the frictional force is negative (-)

d) Work done by the Normal force

W = (N) *d *cos(90°)

W = (101.8 )*(19.9) (0) (N*m)

W = 0

4 0

2 years ago