What's the minimum number of calories to eat per day?

Which sub-atomic particle is positively charged and found in the nucleus?

IrinaK [193]

The proton. There are three main types of subatomic particles. Protons, Neutrons, and Electrons. Protons are positively charged and found in the nucleus. Neutrons are not charged at all, and found in the nucleus, and electrons are negatively charged and found outside the nucleus in regions known as orbitals.

4 0

3 years ago

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A block of mass 0.490 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T

Dennis_Churaev [7]

Answer:

The value of x is $x = 0.1955\ m$

The value of the velocity at the top is $v_{top} = 2.25 \ m/s$

Explanation:

The diagram of this process is on the first uploaded image

From the question we are told that

The mass of the block is $m_b = 0.490kg$

The distance of compression is $x$

The force constant is $k = 450 N/m$

The radius of the circular track is $R =1.00m$

The speed at the bottom of the track is $v_A = 13.4 \ m/s$

The frictional force experienced is $F_f = 7.00 \ N$

Now looking at this process we see that the potential energy of the spring is been transformed into the kinetic energy of the block . So,

$PE \ of \ spring = KE \ of \ block$

mathematically i.e

$\frac{1}{2}k x^2 = \frac{1}{2} m_bv^2_A$

$0.5 * 450 * x^2 = 0.5 * 0.490 * 13.4^2$

Making x the subject of the formula

$x = \sqrt{\frac{0.5 * 0.490 * 13.4^2}{ 0.5 * 450} }$

$= 0.1955\ m$

The average workdone by friction is $W_f = F_f * \pi$

Here $\pi is \ the\ net\ displacement$

$W_f = 7 * 3.142$

$= 21.99 J$

The kinetic energy at the bottom is

$KE = \frac{1}{2} * 0.490 * 13.4^2$

$= 43.99 \ N$

The potential energy gained at the top of the circle is

$PE = m_bgh$

Here h is the height which is equal to d(diameter) = 2r = 2 × 1 = 2 m and

g is acceleration due to gravity $= 9.8m/s^2$

Now substituting values

$PE = 0.490 * 9.8 *2$

$=9.604 J$

Since energy is can not be created nor destroyed but transformed according to first law of thermodynamics

$KE_{at \bottom} = PE _ {\ at \ top } + W_f + KE_{\ at \ top}$

$43.99 = 9.604 +21.99 + [\frac{1}{2} m_b * v_{top} ^2]$

$12.369= [2.45* v_{top} ^2]$

$v_{top} =\sqrt{\frac{12.369}{2.45} }$

$= 2.25 \ m/s$

8 0

3 years ago

On a rectangle with length 100 m, width 50m and 2 vehicles stationed together. Find the time that they meet given that car A tra

zhannawk [14.2K]

Answer:

$T=35.625sec$

Explanation:

From the question we are told that:

Length $L=100 m$

Width $W=50m$

Velocity of Car A $V_A=5m/s$

Velocity of Car B $V_B=3m/s$

Distance traveled by car A before car B moves

$d_l=5*3$

$d_l=15$

Therefore total distance traveled at same time interval

$D=total\ distance-d_l$

Where

Total distance=Perimeter of rectangle

$P=2(L+B)$

$P=2(100+50)$

$P=300$

Therefore

$D=total\ distance-d_l$

$D=300-15\\D=285m$

Generally the equation for time taken to meet is mathematically given by

$T=\frac{Distance D}{Relative\ speed V_r}$

Where

Relative speed = Speed of car A +Speed of car B

$V_r=V_A+V_B$

$V_r=5+3$

$V_r=8m/s$

Therefore the time taken to meet

$T=\frac{ D}{ V_r}$

$T=\frac{ 285}{ 8}$

$T=35.625sec$

8 0

3 years ago

A 200 g block is pressed against a spring of force constant 1.40 kN/m until the block compresses the spring 10.0 cm. The spring

tino4ka555 [31]

Answer:

Explanation:

This problem bothers on the energy stored in a spring in relation to conservation of energy

Given data

Mass of block m =200g

To kg= 200/1000= 0.2kg

Spring constant k = 1.4kN/m

=1400N/m

Compression x= 10cm

In meter x=10/100 = 0.1m

Using energy considerations or energy conservation principles

The potential energy stored in the spring equals the kinetic energy with which the block move away from the spring

Potential Energy stored in spring

P.E=1/2kx^2

Kinetic energy of the block

K.E =1/mv^2

Where v = velocity of the block

K.E=P.E (energy consideration)

1/2kx^2=1/mv^2

Kx^2= mv^2

Solving for v we have

v^2= (kx^2)/m

v^2= (1400*0.1^2)/0.2

v^2= (14)/0.2

v^2= 70

v= √70

v= 8.36m/s

a. Distance moved if the ramp exerts no force on the block

Is

S= v^2/2gsinθ

Assuming g= 9. 81m/s^2

S= (8.36)^2/2*9.81*sin60

S= 69.88/19.62*0.866

S= 69.88/16.99

S= 4.11m

6 0

3 years ago

Which kinds of objects emit light?

Luda [366]

A: objects that shine
Visible: Our eyes detect visible light. Fireflies, light bulbs, and stars all emit visible light. Ultraviolet: Ultraviolet radiation is emitted by the Sun and are the reason skin tans and burns. "Hot" objects in space emit UV radiation as well.

5 0

3 years ago

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