4. Find the density of 2750 g of a substance that occupies 250 mL.

The roller-coaster car shown in fig. 6-41 (h1 = 30 m, h2 = 12 m, h3 = 20 m), is dragged up to point 1 where it is released from

maxonik [38]

<span>Since there is no friction, conservation of energy gives change in energy is zero Change in energy = 0 Change in KE + Change in PE = 0 1/2 x m x (vf^2 - vi^2) + m x g x (hf-hi) = 0 1/2 x (vf^2 - vi^2) + g x (hf-hi) = 0 (vf^2 - vi^2) = 2 x g x (hi - hf) Since it starts from rest vi = 0 Vf = squareroot of (2 x g x (hi - hf)) For h1, no hf Vf = squareroot of (2 x g x (hi - hf)) Vf = squareroot of (2 x 9.81 x 30) Vf = squareroot of 588.6 Vf = 24.26 For h2 Vf = squareroot of (2 x 9.81 x (30 â€“ 12)) Vf = squareroot of (9.81 x 36) Vf = squareroot of 353.16 Vf = 18.79 For h3 Vf = squareroot of (2 x 9.81 x (30 â€“ 20)) Vf = squareroot of (20 x 9.81) Vf = 18.79</span>

7 0

3 years ago

Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k

Harrizon [31]

Explanation:

Given that,

(a) Speed, $v=6.66\times 10^6\ m/s$

Mass of the electron, $m_e=9.11\times 10^{-31}\ kg$

Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

The wavelength of the electron is given by :

$\lambda_e=\dfrac{h}{m_ev}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}$

$\lambda_e=1.09\times 10^{-10}\ m$

The wavelength of the proton is given by :

$\lambda_p=\dfrac{h}{m_p v}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}$

$\lambda_p=5.96\times 10^{-14}\ m$

(b) Kinetic energy, $K=1.71\times 10^{-15}\ J$

The relation between the kinetic energy and the wavelength is given by :

$\lambda_e=\dfrac{h}{\sqrt{2m_eK}}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}$

$\lambda_e=1.18\times 10^{-11}\ m$

$\lambda_p=\dfrac{h}{\sqrt{2m_pK}}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}$

$\lambda_p=2.77\times 10^{-13}\ m$

Hence, this is the required solution.

6 0

3 years ago

A pendulum is made by attaching a sphere to the end of a string of negligible mass and it oscillates when released from an angle

Rudik [331]

Answer:

Angular momentum conservation and kinetic energy. Torsional ... motion-observation of what a given object does in relation to other objects. Frames of ... shows that the rectangular and spherical polar coordinates are related as follows: ... 2mo, which are connected by a string over a pulley of negligible mass and prevented.

Explanation:

8 0

3 years ago

Suppose there is a uniform magnetic field, B, pointing into the page (so your index finger will point into the page). If the vel

Naily [24]

Answer:

Explanation:

We shall show all given data in vector form and calculate the direction of force with the help of following formula

force F = q ( v x B )

q is charge , v is velocity and B is magnetic field.

Given B = - Bk ( i is right , j is upwards and k is straight up the page )

v = v j

F = q ( vj x - Bk )

= -Bqvi

The direction is towards left .

a ) If velocity is down

v = - v j

F = q ( - vj x - bk )

= qvB i

Direction is right .

b ) v = v i

F = q ( vi x - Bk )

= qvB j

force is upwards

c ) v = - vi

F = q ( -vi x - Bk )

= -qvBj

force is downwards

d ) v = - v k

F = q( - vk x -Bk )

= 0

No force will be created

e ) v = v k

F = q( vk x -Bk )

= 0

No force will be created

3 0

3 years ago

20. Consider a model steel bridge that is 1/100 the exact scale of the real bridge that is to be built. a. If the model bridge w

Veseljchak [2.6K]

The model bridge captures all the structural attributes of the real bridge, at a reduced scale.

Part a.
Note that volume is proportional to the cube of length. Therefore the actual bridge will have 100^3 = 10^6 times the mass of the model bridge.

Because the model bridge weighs 50 N, the real bridge weighs
(50 N)*10^6 = 50 MN.

Part b.
The model bridge matches the structural characteristics of the actual bridge.
Therefore the real bridge will not sag either.

6 0

4 years ago

4. Find the density of 2750 g of a substance that occupies 250 mL.​

4. Find the density of 2750 g of a substance that occupies 250 mL.