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2 years ago

Gina estimates that her moms truck weighs about 6,000 pounds. How can you write 6,000 using exponents?

Physics

2 answers:

Georgia [21]2 years ago

5 0

6x10^3 six times ten to the third power :)

Nookie1986 [14]2 years ago

5 0

Answer:

$6\times 10^{3} Pounds$

Explanation:

Given that the Gina moms truck weight is 6000 pounds.

And we also know that the 6000 pounds can be written as in the form.

$6000=6\times 10\times 10\times 10$

Further more it can be written as,

$6000Pounds=6\times 10^{3} Pounds$

Therefore the weight of the truck which is 6000 Pounds can be written in exponential form as $6\times 10^{3} Pounds$ .

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Explain why it is better to jump and land with bent knees rather than straight legs? Use key physics vocabulary.

BigorU [14]

Answer:

Bent knees

Explanation:

Depending the height your jumping from, if you were jumping from a high structure it's best if you jump and land on knees so you dont sprang your ankles from jumping and landing on your straight legs, you could even pull a muscle if you land on straight legs.

3 0

2 years ago

In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,

Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B $=\ 1.0\ Kg.$

Let mass of ball $A$ and $B\ is\ m$

Final velocity of ball $A\ is\ v_1$

Final velocity of ball $B\ is\ v_2$

initial velocity of ball $A\ is\ u_1$

Initial velocity of ball $B\ is\ u_2$

Momentum after collision $=mv_1+mv_2$

Momentum before collision $= mu_1+mu_2$

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, $mu_1+mu_2=mv_1+mv_2$

Plugging each trial in this equation we get,

First Trial

$mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3$

momentum before collision $\neq$ moment after collision

Second Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1$

moment before collision $=$ moment after collision

Third Trial

$mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1$

momentum before collision $\neq$ moment after collision

Fourth Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0$

momentum before collision $\neq$ moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0

3 years ago

Read 2 more answers

Earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.

kobusy [5.1K]

The Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x $10^{24}$ Kg

<h3>Relationship between Linear and angular speed</h3>

Linear speed is the product of angular speed and the maximum displacement of the particle. That is,

V = Wr

Where

V = Linear speed

W = Angular speed

r = Radius

Given that the earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.

a) To determine the Earth’s average orbital speed, we will make use of the below formula to calculate angular speed

W = 2 $\pi$ /T

W = (2 x 3.143) / (365.26 x 24)

W = 6.283 / 876624

W = 7.2 x $10^{-4}$ Rad/hr

The Earth’s average orbital speed V = Wr

V = 7.2 x $10^{-4}$ x 149.6 x $10^{6}$

V = 107225.5 kilometers per hours.

b) Based on the information given in this question, to calculate the approximate mass of the Sun, we will use Kepler's 3rd law

M = (4 $\pi ^{2}$ $r^{3}$ ) / G $T^{2}$

M = (4 x 9.8696 x 3.35 x $10^{24}$ ) / (6.67 x $10^{-11}$ x 7.68 x $10^{11}$ )

M = 1.32 x $10^{26}$ / 51.226

M = 2.58 x $10^{24}$ Kg

Therefore, the Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x $10^{24}$ Kg

Learn more about Orbital Speed here: brainly.com/question/22247460

#SPJ1

3 0

1 year ago

An electron and a proton have the same kinetic energy upon entering a region of constant magnetic field and their velocity vecto

kupik [55]

Answer: rp/re= me/mp= 544 * 10^-6.

Explanation: To calculate this problem we have to consider the circular movement by the electron and proton inside a magnetic field.

Then the dynamic equation for the circular movement is given by:

Fcentripetal= m*ω^2.r

q*v*B=m*ω^2.r

we write this for each particle then we have the following:

q*v*B=me* ω^2*re

q*v*B=mp* ω^2*rp

rp/re=me/mp=9.1*10^-31/1.67*10^-27=544*10^-6

4 0

3 years ago

The decrease of PE for a freely falling object equals its gain in KE, in accord with the conservation of energy. By simple algeb

Delvig [45]

Answer:

$v = \sqrt{20h}$

Explanation:

The potential energy (PE) we are looking here is gravitational potential energy (GPE).

GPE= mgh,

where m is the mass of an object,

g is the gravitational field strength

h is the height of the object

KE= ½mv²,

where m is the mass and v is the velocity

loss in GPE= gain in KE

mgh= ½mv²

gh= ½v² (divide by m throughout)

Assuming that the object is on earth, then g= 10N/Kg

½v²= 10h (substitute g=10)

v²= 20h (×2 on both sides)

v= √20h (square root both sides)

4 0

2 years ago