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BaLLatris [955]

3 years ago

6

Acceleration increases over time once a force is applied to the object. Determine the acceleration at 3.5 sec? A) 3 m/s2 B) 6 m/

s2 C) 15 m/s2 D) 53 m/s2

2 answers:

borishaifa [10]3 years ago

6 0

Answer:

B) 6

Explanation:

I just got it wrong so u could get it right

Novay_Z [31]3 years ago

4 0

Answer: Acceleration does not increase over time once a force is applied to the object. It depends on the force and the object's mass. If those don't change, then the acceleration is constant.

Explanation:. F = M • a

We don't know either of those numbers, so we can't answer the question.

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During which time does Jamal have the greatest momentum?

Yuliya22 [10]

Answer:

Momentum Packet Answer KEY - Science Online

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4 0

3 years ago

There are two different alleles for the number of fingers on the hands: 5 finger allele and 6 finger allele. When both the 5 fin

padilas [110]

Answer:

The 6 fingers allele is dominant

Explanation:

We are told that the the individual is genotypically heterozygous, that is the have both types of the finger allele: the 5 finger allele and the 6 fingers allele however phenotypically, 6 fingers are observed. From this we can conclude that the 6 fingers allele is the one that is dominant because it is the one that is expressed phenotypically.

8 0

3 years ago

Recall from Chapter 1 that a watt is a unit of en- ergy per unit time, and one watt (W) is equal to one joule per second ( J·s–1

harkovskaia [24]

Answer:

Explanation:

The energy of a photon is given by the equation $E_p=h f$ , where h is the <em>Planck constant</em> and f the frequency of the photon. Thus, N photons of frequency f will give an energy of $E_N=N h f$ .

We also know that frequency and wavelength are related by $f=\frac{c}{\lambda}$ , so we have $E_N=\frac{N h c}{\lambda}$ , where c is the <em>speed of light</em>.

We will want the number of photons, so we can write

$N=\frac{\lambda E_N}{h c}$

We need to know then how much energy do we have to calculate N. The equation of power is $P=E/t$ , so for the power we have and considering 1 second we can calculate the total energy, and then only consider the 4% of it which will produce light, or better said, the N photons, which means it will be $E_N$ .

Putting this paragraph in equations:

$E_N=(\frac{4}{100})E=0.04Pt=(0.04)(100W)(1s)=4J$ .

And then we can substitute everything in our equation for number of photons, in S.I. and getting the values of constants from tables:

$N=\frac{\lambda E_N}{h c}=\frac{(520 \times10^{-9}m) (4J)}{(6.626\times10^{-34}Js) (299792458m/s)}=1.047 \times10^{19}$

3 0

3 years ago

A key falls from a bridge that is 45 m above the water. the key falls straight down and lands in a model boat traveling at a con

erastova [34]

Let the key is free falling, therefore from equation of motion

$h = ut +\frac{1}{2}gt^2.$ .

Take initial velocity, u=0, so

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2$ .

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }$

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

$d= v \times t$

From above substituting t,

$d = v \times \sqrt{\frac{2h}{g} }$ .

Now substituting all the given values and g = 9.8 m/s^2, we get

$d = 3.5 \ m/s \times \sqrt{\frac{2 \times 45 m}{9.8 m/s^2} } = 10.60 m$ .

Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.

7 0

3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago