if one paperclip has the mass of 1 gram and 1000 paperclips has a mass of 1 kilogram how many kilograms are 8000 paperclips ?

A farm truck travels due east with a constant speed of 9.50 m/s along a horizontal road. A boy riding in the back of the truck t

Reil [10]

Answer:

a)he angle is 90 with respect to the horizontal (x axis)

b) its speed is zero both vertically and horizontally

c) vertical path

d) a parabolic movement

e) v₀ = (9.5i ^ + 8.232 j ^) m / s

Explanation:

This is a relative problem and movement.

.v ’= v + u

Where v ’is the speed with respect to the mobile system, v the speed with respect to the fixed system and the speed between the two reference systems

a) The child and the can is in the truck, so they go at the speed of the truck, when he throws the can he continues at this speed on the x-axis and therefore as the two advance the same distance the more hands of the child, consequently the can is thrown vertically

The angle is 90 with respect to the horizontal (x axis)

b) with respect to the truck, the can is still, so its speed is zero both vertically and horizontally

c) The child sees that the can follows a vertical path

d) A stationary observer on the ground, sees that the can has a constant speed in the same direction of the truck and when they throw it vertical goal has a vertical movement, the sum of these two movements gives a parabolic movement of the same uncle as a projectile launch

e) the initial speed has two components

X Axis v_lata = v_camion = 9.5 m / s

Y Axis speed given by the child

Let's look for the travel time

x = v₀ₓ t

t = x / v₀ₓ

t = 16 / 9.5

t = 1.68 s

y = $v_{oy}$ t - ½ g t²

When he returns to the child's hand and = 0

0 = v_{oy} t - ½ g t²

v_{oy} = ½ g t

v_{oy} = ½ 9.8 1.68

v_{oy} = 8,232 m / s

Speed is

v₀ = (9.5i ^ + 8.232 j ^) m / s

5 0

3 years ago

What occurs at the end of a negative cell if an electric cell is placed in a closed circuit?

gavmur [86]

1. Current means flow of electrons and electrons are negative charged and so are attracted to the positive end of the battery and repelled by the negative end.

2. When battery is connected in a circuit that lets the electron flow through it, they flow from negative to positive.

3. When negative terminal of cell is connected to other negative terminal of the cell in a particular circuit then, current will not flow in circuit as electrons cannot flow from negative to negative terminal.

4 0

3 years ago

A 8.10×10^3 ‑kg car is travelling at 25.8 m/s when the driver decides to exit the freeway by going up a ramp. After coasting 3.9

KengaRu [80]

Answer:

The magnitude of the average drag force is 2412.34 N.

Explanation:

Given that,

Mass of car $m=8.10\times10^{-3}\ kg$

Velocity v = 25.8 m/s

Distance $d= 3.90\times10^{2}$

Speed of car = 13.1 m/s

Height = 12.5 m

We need to calculate the magnitude of the average drag force

Using equation kinetic energy

$K.E_{i}=K.E_{f}+P.E+F_{d}$

$\dfrac{1}{2}mv_{i}^2=\dfrac{}{}mv_{f}^2+mgh+F\times d$

Where, $v_{i}$ = initial velocity

$v_{f}$ = final velocity

h = height

g = acceleration due to gravity

$F_{d}$ =drag force

m = mass of the car

d = distance

Put the value into the formula

$\dfrac{1}{2}\times8.10\times10^{3}\times25.8=\dfrac{1}{2}\times8.10\times10^{3}\times13.1+8.10\times10^{3}\times9.8\times12.5+F\times3.90\times10^{2}$

$F=\dfrac{\dfrac{1}{2}\times8.10\times10^{3}\times25.8-\dfrac{1}{2}\times8.10\times10^{3}\times13.1-8.10\times10^{3}\times9.8\times12.5}{3.90\times10^{2}}$

$F=-2412.34\ N$

$|F|=2412.34\ N$

Hence, The magnitude of the average drag force is 2412.34 N.

4 0

4 years ago

Let k be the Boltzmann constant. If the configuration of the molecules in a gas changes so that the multiplicity is reduced to o

katen-ka-za [31]

Answer:

ΔS = - k ln (3)

Explanation:

Using the Boltzmann's expression of entropy, we have;

S = k ln Ω

Where;

S = Entropy

Ω = Multiplicity

From the question, the configuration of the molecules in a gas changes so that the multiplicity is reduced to one-third its previous value. This also causes a change in the entropy of the gas as follows;

ΔS = k ln (ΔΩ)

ΔS = kln(Ω₂) - kln(Ω₁)

ΔS = kln(Ω₂ / Ω₁) -------------(i)

Where;

Ω₂ = Final/Current value of the multiplicity

Ω₁ = Initial/Previous value of the multiplicity

Ω₂ = $\frac{1}{3}$ Ω₁ [since the multiplicity is reduced to one-third of the previous value]

Substitute these values into equation (i) as follows;

ΔS = k ln ( $\frac{1}{3}$ Ω₁ / Ω₁)

ΔS = k ln ( $\frac{1}{3}$ )

ΔS = k ln (3⁻¹)

ΔS = - k ln (3)

Therefore, the entropy changes by - k ln (3)

4 0

3 years ago

A hiker leaves her camp and walks 3.5 km in a direction of 55° south of west to the lake. After a short rest at the lake, she hi

7nadin3 [17]

1) Magnitude

Let's take south as positive y-direction and east as positive x-direction. Then we have to resolve both displacements into their respective components:

$d_{1x} = -(3.5 km) cos 55^{\circ}=-2.0 km$