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3 years ago

VU

Physics

2 answers:

aliya0001 [1]3 years ago

7 0

Answer:

1350 meters

Explanation:

45x30=1350

Serga [27]3 years ago

6 0

Answer:

An ant can move 1350 meters in 45 minutes.

Explanation:

The average velocity is defined as:

$v = \frac{d}{t}$ (1)

Where v is the velocity, d is the distance and t is the time

Then, d can be isolated from equation 1:

$d = v \cdot t$ (2)

The velocity of the ant is 30 m/min (30 meters per minute). Which means that the ant can cover 30 meters in one minute.

Finally, equation 2 can be used.

$d = (30 \frac{m}{min}) \cdot (45min)$

$d = 1350m$

Hence, an ant can move 1350 meters in 45 minutes.

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At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/

Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

4 0

3 years ago

Scientific theories can change over time as new information is discovered if a scientififc theory changes does this mean that it

Verdich [7]

Not necessarily, it just means that the Scientific theory was not complete and needs additional information, research, and ideas.

3 0

3 years ago

Read 2 more answers

If ball C is 3 times the volume of ball D and ball D has 1/3 the mass of ball C, which has the greater density?

Alex Ar [27]

Answer:

They have same density

Explanation:

The density of an object is defined as

$d=\frac{m}{V}$

where

m is the mass of the object

V is its volume

Let's call $m_c$ and $V_c$ the mass and the volume of ball C, respectively. Therefore, the density of ball C is:

$d_c = \frac{m_c}{V_c}$

We know that the volume of ball C is 3 times the volume of ball D, so

$V_c = 3 V_d \rightarrow V_d = \frac{V_c}{3}$

And we also know that ball D has 1/3 the mass of ball C:

$m_d = \frac{m_c}{3}$

So, the density of ball D is:

$v_d = \frac{m_d}{V_d}=\frac{m_c/3}{V_d/3}=\frac{m_c}{V_c}=d_c$

Therefore, the two balls have same density.

6 0

4 years ago

An electron has a kinetic energy of 10.1 eV. The electron is incident upon a rectangular barrier of height 18.2 eV and width 1.0

podryga [215]

Answer:

factor that the electron's probability of tunneling through the barrier increase 2.02029

Explanation:

given data

kinetic energy = 10.1 eV

height = 18.2 eV

width = 1.00 nm

wavelength = 546 nm

solution

we know that probability of tunneling is express as

probability of tunneling = $e^{-2CL}$ .................1

here C is = $\frac{\sqrt{2m(U-E}}{h}$

here h is Planck's constant

c = $\frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-10.1) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}$

c = 2319130863.06

and proton have hf = $\frac{hc}{\lambda } = {1240}{546}$ = 2.27 ev

so electron K.E = 10.1 + 2.27

KE = 12.37 eV

so decay coefficient inside barrier is

c' = $\frac{\sqrt{2m(U-E}}{h}$

c' = $\frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-12.37) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}$

c' = 1967510340

the factor of incerease in transmisson probability is

probability = $e^{2L(c-c')}$

probability = $e^{2\times 1\times 10^{-9} \times (351620523.06)}$

factor probability = 2.02029

3 0

3 years ago

A 1.89 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.77 N

Maksim231197 [3]

Answer:

Magnitude F(t)=26.6 N

Direction: -x

Explanation:

Given data

Spring constant K=4.77 N/m

Mass m=1.89 kg

Displace A=5.56m

Time t=3.96s

To find

Magnitude of force F

Solution

The angular frequency is given as

$w=\sqrt{\frac{K}{m} } \\w=\sqrt{\frac{4.77N/m}{1.89kg} }\\w=1.59rad/s$

Force on object is

$F(t)=-mAw^{2}Cos(wt)$

Substitute given values

$F(t)=-(1.89kg)(5.56m)(1.59rad/s)^{2}Cos(1.59*3.96)\\F(t)=-26.6N$

Magnitude F(t)=26.6 N

Direction: -x

4 0

3 years ago