Ask question

Ask question

All categories

2 years ago

14

Select all of the answers that apply. Which of the following make the Earth unique when compared to the other planets in our sol

ar system?

1 answer:

zalisa [80]2 years ago

7 0

Earth is unique because Earth has gravity and other planets don't.

You might be interested in

When using a micron gauge, if the pressure gauge falls frim 1,300 to 1,000 instantly, the vehicle needs to?

julsineya [31]

If the micron gauge pressure falls instantly, it means that the moisture of the air-conditioning is frozen, therefore you need to move the vehicle to a warmer place in order to unfreeze the moisture.

5 0

3 years ago

A 1560-kilogram truck moving with a speed of 28.0 m/s runs into the rear end of a 1070-kilogram stationary car. If the collision

Nimfa-mama [501]

Answer:

Δ KE = 249158.6 kJ

Explanation:

given data

Truck mass M = 1560 Kg

Truck initial speed, u = 28 m/s

mass of car m = 1070 Kg

initial speed of car u1 = 0 m/s

solution

first we get here final speed by using conservation of momentum that is express as

Mu = (M+m) V .......................1

put here value we get

1560 × 28 = (1560 + 1070 ) V

solve it we get

final speed V = 16.60 m/s

and

Change in kinetic energy will be here

Δ KE = $\frac{1}{2} Mu^2 - \frac{1}{2}(M+m)V^2$ .................2

put here value and we get

Δ KE = $\frac{1}{2}\times 1560\times 28^2 - \frac{1}{2}\times (1560 + 1070)\times 16.60^2$

solve it we get

Δ KE = 249158.6 kJ

6 0

3 years ago

The nucleus of an atom can have either a positive or negative charge. please select the best answer from the choices provided t

Airida [17]

F - False.

The nucleus of an atom is positively charge.

5 0

3 years ago

Read 2 more answers

Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre

castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:

V

=

d

x

/d

t

=

7.5

×m

i

/

h

r

...........................(

1

)

in Above equation in x and t. Separating the variables and integrating,

∫

d

x

/7.5

×=

∫

d

t

+

C

or

−

4.7619

=

t

+

C

Here C =constant of integration.

x

=

0 at t

=

0

, we get: C

=

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619 =

t

−

4.7619

.

.

.

.

.

.

.

.

.

(

2

)

Here

x is measured in miles and t in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),

to get:

= −

4.7619

=

1

−

4.7619

= −

3.7619

or x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in m

i

l

/

differentiate

equation (1) with respect to time.

we have to eliminate x from the equation (1) using equation (2).

Eliminating x we get:

v

=

7.5×

Now differentiating above equation w.r.t time we get:

a

=

d

v/

d

t

=

−

0.675

/

At

t

=

0

the joggers acceleration is :

a

=

−

0.675

m

i

l

/

=

−

4.34

×

f

t

/

(c) required time for the jogger to run 6 miles is obtained by setting

x

=

6 in equation (2). We get:

−

4.7619

(

1

−

(

0.04

×

6 )

)^

7

/

10=

t

−

4.7619

or

t

=

0.832

h

r

s

6 0

3 years ago

A train travels due north in a straight line with a constant speed of 100 m/s. Another train leaves a station 2,881 m away trave

damaskus [11]

Answer:

The trains will collide at a distance 1660 m from the station

Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

∴ If train A has covered a distance $x$ m by the time of collision, then train B would have traveled $(2881 - x)$ m.

Also,

At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

$t_{A} = t_{B}$

$t_{A}$ is the time spent by train A

$t_{B}$ is the time spent by train B

From,

$Velocity = \frac{Distance }{Time }\\$

$Time = \frac{Distance}{Velocity}$

Since the time spent by the two trains is equal,

Then,

$\frac{Distance_{A} }{Velocity_{A} } = \frac{Distance_{B} }{Velocity_{B} }$

${Distance_{A} = x$ m

${Distance_{B} = 2881 - x$ m

${Velocity_{A} = 100$ m/s

${Velocity_{B} = 136$ m/s

Hence,

$\frac{x}{100} = \frac{2881 - x}{136}$

$136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\$

$x$ ≅ 1,221 m

This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

7 0

3 years ago