Ask question

Ask question

All categories

3 years ago

5

CAN PEOPLE PLEASE HELP, I NEED TO GET TO THE NEXT LEVEL OR WHATEVER ITS CALLED BUT I NEED 5 BRAINLIEST IN 2 DAYS, I ALREADY HAVE

7 BUT THAT DOSENT COUNT IG, IF YOU HELP ILL RETURN THE FAVOR AND GIVE YOU BRAINLIEST ON YOUR AWENSERS OR QUESTIONS........ ILL AWENSER ONE OF YOUR QUESTIONS AND YOU MAKE ME BRAINLIEST? THXX

2 answers:

Vilka [71]3 years ago

8 0

Answer:

yes you can go on your account

lapo4ka [179]3 years ago

6 0

Answer:

answer the question i have asked go to my account

Explanation:

can i get brainliest

You might be interested in

Two wheels having the same radius and mass rotate at the same angular velocity. One wheel is made with spokes so nearly all the

Mazyrski [523]

Answer:

C. The wheel with spokes has about twice the KE.

Explanation:

Given that

Mass , radius and the angular speed for both the wheels are same.

radius = r

Mass = m

Angular speed = ω

The angular kinetic energy KE given as

$KE=\dfrac{1}{2}I\omega ^2$

I=Moment of inertia for wheels

Wheel made of spokes

I₁ = m r²

Wheel like a disk

I₂ = 0.5 m r²

Now by comparing kinetic energy

$\dfrac{KE_1}{KE_2}=\dfrac{I_1}{I_2}$

$\dfrac{KE_1}{KE_2}=\dfrac{mr^2}{0.5mr^2}$

$\dfrac{KE_1}{KE_2}=2$

KE₁= 2 KE₂

Therefore answer is C.

5 0

3 years ago

Which is an example of a physical change?

Ber [7]

Answer:

salt dissolving in water

6 0

3 years ago

Read 2 more answers

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b

Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy= $\frac{1}{2} kx^{2} +PE1$

Energy at the point where the block will stop consists of only gravitational potential energy= $PE2$

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒ $\frac{1}{2} kx^{2} +PE1=PE2$

⇒ $PE2-PE1=\frac{1}{2} kx^{2}$

Also $PE2-PE2=mgh$

where $m$ is the mass of block

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

$h$ is the difference in height between two positions

⇒ $mgh=\frac{1}{2} kx^{2}$

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴ $2100*9.8*h=\frac{1}{2}*2200*0.11^{2}$

⇒ $20580*h=13.31$

⇒ $h=\frac{13.31}{20580}$

⇒h=0.0006467m= $6.5e-4$

7 0

4 years ago

6- A metal block measures 10 cm x 2 cm x 2 cm. what is its volume? how many blocks each

zalisa [80]

You need 5 blocks of the smaller object to contain the same amount of volume of the bigger object

3 0

4 years ago

A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has

iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
So, we can write the following general equation, taking the initial and final values of the energies:

$\Delta K + \Delta U = W_{ffr} (1)$

Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
⇒ Kf = 1/2*m*vf² (2)
The change in the potential energy, can be written as follows:

$\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)$

where k = force constant = 815 N/m

xf = final displacement of the block = 0.01 m (taking as x=0 the position

for the spring at equilibrium)

x₀ = initial displacement of the block = 0.03 m

Regarding the work done by the force of friction, it can be written as follows:

$W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)$

where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

horizontal displacement.

Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
Fn = Fg= m*g (5)
Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

$\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)$

$\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)$

Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

$\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)$

$v =\sqrt{(0.326-0.1568} = 0.41 m/s (9)$

7 0

3 years ago