Find the acceleration of a forklift of mass 1400 kg pushed by a force of 2100 N that is opposed by friction force of 425 N

We observe that a moving charged particle experiences no magnetic force. From this we can definitely conclude that:_______

Leto [7]

Answer:

b. the particle must be moving parallel to the magnetic field.

Explanation:

The magnetic force on a moving charged particle is given by;

F = qvBsinθ

where;

q is the charge of the particle

v is the velocity of the particle

B is the magnetic field

θ is the angle between the magnetic field and velocity of the moving particle.

When is the charge is stationary the magnetic force on the charge is zero.

Also when the charge is moving parallel to the magnetic field, the magnetic force is zero.

Therefore, when a moving charged particle experiences no magnetic force, we can definitely conclude that the particle must be moving parallel to the magnetic field.

b. the particle must be moving parallel to the magnetic field.

5 0

2 years ago

The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum

Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.

5 0

2 years ago

If 56 grams of carbon monoxide burns in oxygen to produce 88 grams of carbon dioxide, the mass of oxygen involved in the reactio

pentagon [3]

Answer : The mass of oxygen involved in the reaction is, 32 grams.

Explanation : Given,

Mass of carbon monoxide = 56 g

Mass of carbon dioxide = 88 g

Molar mass of carbon monoxide $(CO)$ = 28 g/mole

Molar mass of oxygen $(O_2)$ = 32 g/mole

First we have to calculate the moles of carbon monoxide.

$\text{Moles of CO}=\frac{\text{Mass of CO}}{\text{Molar mass of CO}}=\frac{56g}{28g/mole}=2moles$

Now we have to calculate the moles of oxygen gas.

The balanced chemical reaction will be,

$2CO+O_2\rightarrow 3CO_2$

From the balanced chemical reaction, we conclude that

2 moles of CO react with 1 mole of $O_2$

So, the moles of $O_2$ = 1 mole

Now we have to calculate the mass of oxygen.

$\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2$

$\text{Mass of }O_2=1mole\times 32g/mole=32g$

Therefore, the mass of oxygen involved in the reaction is, 32 grams.

5 0

2 years ago

A hunter wishes to cross a river that is 1.5 km wide and flows with a velocity of 5.0 km/h parallel to its banks. The hunter use

anzhelika [568]

Answer:137.48 s

Explanation:

Given

Width of river=1.5 km

velocity of river=5 km/h

velocity of boat w.r.t river =12 km/h

To cross the river in minimum time hunter needs to cross the river perpendicular to the flow

i.e. velocity of boat w.r.t water must be perpendicular

i.e. x component of boat must be equal to river flow

$12cos\theta =5$

where $\theta$ is angle made by boat w.r.t bank

$cos\theta =\frac{5}{12}$

$cos\theta =0.416$

$\theta =65.417^{\circ}$

its vertical component is $12sin(65.417)=10.91 m/s$

Time taken $=\frac{1.5\times 1000}{10.91}=137.48 s$

6 0

2 years ago

A ballistic pendulum is a device used to measure the speed of a bullet. A bullet is fired at a block of wood hanging from two st

laila [671]

Answer:

1.56 m/s, 0.124 m

Explanation:

mass of bullet, m = 6.5 g = 0.0065 kg

initial velocity of bullet, u = 530 m/s

mass of block, M = 2.2 kg

initial velocity of block, v = 0 m/s

Let the speed of bullet and block system after the collision is V.

By use of conservation of momentum

Momentum of system before collision = Momentum of system after collision

Momentum of bullet before collision + momentum of block before collision = Momentum of (bullet + block) after collision

m x u + M x v = (M + m) x V

0.0065 x 530 + 0 = (2.2 + 0.0065) x V

3.445 = 2.2065 x V

V = 1.56 m / s

Thus, the velocity of bullet and block system after collision is 1.56 m/s.

Now use the conservation of energy

The kinetic energy at the bottom is equal to the potential energy at the height.

Let the height raised is h.

$\frac{1}{2}(M+m)V^{2}=(M+m)gh$

$\frac{1}{2}\times 1.56 \times 1.56=9.8\times h$

h = 0.124 m

Thus, the height raised is 0.124 m.

6 0

2 years ago