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horsena [70]
2 years ago
13

A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl

ane by an ideal spring that is aligned with the surface and attached to a wall above the brick.
The spring has a spring constant (force constant) of 120 N/m.

By how much does the spring stretch with the brick attached?- 240 cm- 24 cm- 14 cm- 36 cm- 360 cm
Physics
1 answer:
Firdavs [7]2 years ago
4 0

Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

m (Mass)=3.0 kg

α(incline) =34°

Spring Constant (force constant)=120 N/m

d (distance)=?

Solution

F=mg

F=(3.0)(9.8)

F=29.4 N

As we also know that

Force parallel to the incline=FSinα

F=29.4×Sin(34)

F=16.44 N

d(distance)=F/Spring Constant

d(distance)=16.44/120

d(distance)=0.137 m ⇒13.7 cm

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A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.90 s. You may ignore air
kozerog [31]

Answer:

h=17m

v=18.6 m/s

Explanation: The question can be solved by applying kinematic equations of motion

Data

u=0

a=g

t=1.9 secs

firstly to calculate the height

s=ut+0.5at^2\\h=ut+0.5at^2\\h=0*1.9+0.5*9.81*1.9^2\\h=17.707 m

to find the final velocity

v=u+at\\v=0+9.81*1.9\\v=18.639

The acceleration graph is straight line of equation y=9.8 as acceleration is constant:

Velocity graph is given by y=9.8x ( y as velocity and x as time):

Displacement graph is given by y=4.9x^2 ( x as time, y as displacement):

These graphs are only applicable from x=0 to x=1.9 ... ignore the other graph sections

5 0
2 years ago
A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision
TEA [102]

Answer:

D) No, since kinetic energy is not conserved.

Explanation:

Since momentum is always conserved in all collision

so in Y direction we can say

0 = m(3.5 sin30) - mv_y

v_y = 1.75 m/s

Now similarly in X direction we will have

m(4.5) = m(3.5 cos30 ) + mv_x

v_x = 1.47 m/s

now final kinetic energy of both puck after collision is given as

KE_f = \frac{1}{2}m(3.5^2) + \frac{1}{2}m(1.75^2 + 1.47^2)

KE_f = 8.73 m

initial kinetic energy of both pucks is given as

KE_i = \frac{1}{2}m(4.5^2) + 0

KE_i = 10.125 m

since KE is decreased here so it must be inelastic collision

D) No, since kinetic energy is not conserved.

5 0
2 years ago
What is the concentration of molecular oxygen (O2) in mol/L on a June day in Toronto when atmospheric pressure is 1.0 atm and th
saveliy_v [14]

Answer:

The concentration of mole evil at oxygen on that day is 0.00858 mol/L

Explanation:

Here, we want to calculate the concentration of molecular oxygen

The pressure on that day is 1.0 atm

Since oxygen is at a concentration of 21%, the pressure of oxygen will be 21/100 * 1 = 0.21 atm

Now let’s calculate the concentration;

From Ideal gas law;

PV = nRT

This can be written as;

P/RT = n/V

The term n/V refers to concentration;

Let’s make substitutions now;

P = pressure = 0.21 atm

R = molar gas constant = 0.0821 L•atm/mol•k

T = temperature = 25 = 25 + 273.15 = 298.15 K

Substituting these values, we have;

n/V = C = 0.21/(0.0821 * 298.15) = 0.00858 mol/L

6 0
2 years ago
What is the total flux φ that now passes through the cylindrical surface? enter a positive number if the net flux leaves the cyl
trasher [3.6K]

Net flux through the cylindrical surface is given as

\phi = \frac{q}{epsilon_0}

here q = enclosed charge in the surface

so here in order to find the value of q

q = \lambda* L

so now we have

\phi = \frac{\lambda * L}{\epsilon_0}

so this is the total flux

now by Gauss's law we can find the electric field

\int E.dA = \phi

\int E.dA = \frac{\lambda * L}{\epsilon_0}

E* 2\pi rL = \frac{\lambda * L}{epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

<em>by above expression we can find the electric field at required position</em>

8 0
1 year ago
Complete the statements using data from Table A of your Student Guide. The speed of the cart after 8 seconds of Low fan speed is
finlep [7]

Answer:

The speed of the cart after 8 seconds of Low fan speed is  72.0 cm/s

The speed of the cart after 3 seconds of Medium fan speed is   36.0 cm/s

The speed of the cart after 6 seconds of High fan speed is  96.0 cm/s

Explanation:

took the test on edgenuity

4 0
2 years ago
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