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4 years ago

What is 2.5 × 104 × 3 × 102?

1 answer:

5 0

I assume these are scientific notation thus 10^x not 10x
first multiply all numbers together
so 2.5*10^4*3*10^2
or 25000*300
so you get 7500000.0
move the decimal to 7.5
which is 6 places
the answer is c 7.5*10^6

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If the mass and force of a cart is doubled would its acceleration double or quadruple? also free points <3

DIA [1.3K]

Double , the force is pushed against the cart added by mass is enough to double the acceleration , and thank u :)

8 0

4 years ago

Read 2 more answers

You are writing a short report to a familiar audience and plan to use an informal

miskamm [114]

The thing that the presenter can do to make the presentation to appear informal is to use contractions such as won't, it's, let's, and we're.

<h3>What is a report?</h3>

The term report refers to any formal presentation. It is usually made as part of an academic course or as part of professional development. It could be given while a person is hoping for promotion at a job that he or she is doing.

We have to bear in mind that the audience to whom the report is being presented is a familiar audience hence the presentation can be quite informal and devoid of the formality of presentation of reports as we know it to be.

Now, the focus is to make the presentation to look informal. The thing that the presenter can do to make the presentation to appear informal is to use contractions such as won't, it's, let's, and we're.

Learn more about a report:brainly.com/question/14969693

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7 0

2 years ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

miss Akunina [59]

Formula of the gravitational force between two particles:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67 \cdot 10^{-11} Nm^2 kg^{-2}$ is the gravitational constant

m1 and m2 are the masses of the two particles

r is their distance

(a) particle A

The gravitational force exerted by particle B on particle A is

$F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N$ to the right

The gravitational force exerted by particle C on particle A is

$F_C=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=5.6 \cdot 10^{-6} N$ to the right

So the net gravitational force on particle A is

$F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N$ to the right

(b) Particle B

The gravitational force exerted by particle A on particle B is

$F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N$ to the left

The gravitational force exerted by particle C on particle B is

$F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N$ to the right

So the net gravitational force on particle B is

$F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N$ to the right

The gravitational force exerted by particle A on particle C is

$F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N$ to the left

The gravitational force exerted by particle B on particle C is

$F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N$ to the left

So the net gravitational force on particle C is

$F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N$ to the left

3 0

3 years ago

Calculate the force between two objects that have masses of 70 kilograms and 2,000 kilograms separated by a distance of 1 meter.

Makovka662 [10]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The Gravitational Force between given objects will be ~

$9.34 \times {10}^{ - 6} \: \: N$

$\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}$

We know that ~

$\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}$

where ~

F = gravitational force

$m_1$ = mass of 1st object = 70 kg

$m_2$ = mass of 2nd object = 2000 kg

G = gravitational constant = $6.674 × {10}^ {-11}$

r = distance between the objects = 1 m

Let's calculate the force ~

$F = \dfrac{6.674 \times 10 {}^{ - 11} \times 70 \times 2000 }{1 {}^{2} }$

$F =6.674 \times 7 \times 2 \times 10 { }^{ - 11} \times 10 {}^{4}$

$93.436 \times 10 {}^{ - 7}$

$9.3436 \times 10 {}^{ - 6} \: \: newtons$

6 0

3 years ago

PLEASEEEE HEEEEELP!!!!!

8090 [49]

Answer:

Before:

$p_{truck}=16400\ kg.m/s$

$p_{car}=10000\ kg.m/s$

After:

$p_{truck}=8000\ kg.m/s$

$p_{car}=8400\ kg.m/s$

$v_{fcar}=8.4\ m/s$

$F=9333.33 \ Nw$

Explanation:

<u>Conservation of Momentum</u>

Two objects of masses m1 and m2 moving at speeds v1o and v2o respectively have a total momentum of

$p_1=m_1v_{1o}+m_2v_{2o}$

After the collision, they have speeds of v1f and v2f and the total momentum is

$p_2=m_1v_{1f}+m_2v_{2f}$

Impulse J is defined as

$J=F.t$

Where F is the average impact force and t is the time it lasted

Also, the impulse is equal to the change of momentum

$J=\Delta p$

As the total momentum is conserved:

$p_1=p_2$

$m_1v_{1o}+m_2v_{2o}=m_1v_{1f}+m_2v_{2f}$

We can compute the speed of the second object by solving the above equation for v2f

$\displaystyle v_{2f}=\frac{ m_1v_{1o}+m_2v_{2o} -m_1v_{1f} }{ m_2 }$

The given data is

$m_1=2000\ kg$

$m_2=1000\ kg\\v_{1o}=8.2\ m/s\\v_{2o}=0\ m/s\\v_{1f}=4\ m/s$

a) The impulse will be computed at the very end of the answer

b) Before the collision

$p_{truck}=2000\cdot 8.2=16400\ kg.m/s$

$p_{car}=1000\cdot 0=0\ kg.m/s$

c) After collision

$p_{truck}=2000\cdot 4=8000\ kg.m/s$

Compute the car's speed:

$\displaystyle v_{2f}=\frac{ 16400+0 -8000 }{ 1000 }$

$v_{2f}=8.4\ m/s$

And the car's momentum is

$p_{car}=1000\cdot 8.4=8400\ kg.m/s$

The Impulse J of the system is zero because the total momentum is conserved, i.e. \Delta p=0.

We can compute the impulse for each object

$J_1=\Delta p_1=2000(4-8.2)=-8400 \N.s$

The force can be computed as

$\displaystyle F=\frac{J}{t}=-\frac{8400}{0.9}=-9333.33 \ Nw$

The force on the car has the same magnitude and opposite sign

7 0

4 years ago