What is the weight of a 20 kg box on the Earth? A) 2 N B) 20 N C) 200 N

If the intensity of light that is incident on a piece of metal is increased, what else will be increased? Choose all that apply.

natka813 [3]

Answer:

explained

Explanation:

When the intensity of light is increased on a piece of metal only the number of electron ejected will increase because all other things independent of intensity of light.

Light below certain frequency will not cause any electron emission no matter how intense.

The intensity produces more electron but does not change the maximum kinetic energy of electrons.

Work function is independent of the intensity of light, because it is an intrinsic property of a material.

7 0

3 years ago

In a discussion person A is talking 1.2 dB louder than person B, and person C is talking 3.2 dB louder than person A. What is th

liberstina [14]

Answer: 3.84dB

Explanation:

Since person A is talking 1.2dB louder than B, we will have

A = 1.2B... (1)

Similarly, person C is talking 3.2 dB louder than person A, we have

C = 3.2A... (2)

From equation 1, B = A/1.2... (3)

To get the ratio of the sound intensity of person C to the sound intensity of person B, we will divide equation 2 by 3 to give

C/B = 3.2A/{A/1.2}

C/B = 3.2A×1.2/A

C/B = 3.2×1.2

C/B = 3.84dB

3 0

3 years ago

A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to

Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

$p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa$

On the lower part of the hatch, there is a pressure equal to

$p_{bot}=p_{atm}=1.013\cdot 10^5 Pa$

So, the net pressure acting on the hatch is

$p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa$

which acts from above.

The area of the hatch is given by:

$A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2$

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

$F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N$

8 0

3 years ago

Inside a 30.2 cm internal diameter stainless steel pan on a gas stove water is being boiled at 1 atm pressure. If the water leve

dybincka [34]

Answer:

Q = 20.22 x 10³ W = 20.22 KW

Explanation:

First we need to find the volume of water dropped.

Volume = V = πr²h

where,

r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m

h = height drop = 1.45 cm = 0.0145 m

Therefore,

V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

m = ρV

where,

m = mass = ?

ρ = density of water = 1000 kg/m³

Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

m = 1.038 kg

Now, we calculate the heat required to vaporize this amount of water.

q = mH

where,

H = Heat of vaporization of water = 22.6 x 10⁵ J/kg

Therefore,

q = (1.038 kg)(22.6 x 10⁵ J/kg)

q = 23.46 x 10⁵ J

Now, for the rate of heat transfer:

Rate of Heat Transfer = Q = q/t

where,

t = time = (18.6 min)(60 s/1 min) = 1116 s

Therefore,

Q = (23.46 x 10⁵ J)/1116 s

Q = 20.22 x 10³ W = 20.22 KW

8 0

3 years ago

(a) A light-rail commuter train accelerates at a rate of 1.35 m/s2 . How long does it take to reach its top speed of 80.0 km/h,

Mashcka [7]

Answer:

(a) Time t = 16.46 sec

(b) Time t =13.466 sec

(c) Deceleration = $2.677m/sec^2$

Explanation:

(a) As the train starts from rest its initial velocity u = 0 m/sec

Acceleration $a=1.35m/sec^2$

Final speed v = 80 km/hr

$80km/hr=\frac{80\times 1000}{3600sec}=22.22m/sec$

From first equation of motion v =u+at

So $t=\frac{v-u}{a}=\frac{22.22-0}{1.35}=16.46 sec$

(b) Now initial speed u = 22.22 m/sec

As finally train comes to rest so final speed v=0 m/sec

Deceleration $a=1.65m/sec^2$

So $t=\frac{v-u}{a}=\frac{0-22.22}{-1.65}=13.466 sec$

(c) We have given that initial velocity = 80 km/hr = 22.22 m/sec

Final velocity v = 0 m/sec

Time t = 8.30 sec

So acceleration is given by

$a=\frac{v-u}{t}=\frac{0-22.22}{8.3}=-2.6771m/sec^2$

As acceleration is negative so it is a deceleration

7 0

4 years ago