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harkovskaia [24]

3 years ago

7

Climates on Earth get _____ as you move from the equator to the poles.

2 answers:

andreev551 [17]3 years ago

5 0

The correct answer that would best complete the given statement above would be the last option: COLDER. Climates on Earth get colder <span>as you move from the equator to the poles. The places that are located near or on the equator experience the warmest or the hottest climates such as Africa. Hope this answer helps. </span>

White raven [17]3 years ago

5 0

Answer:

C. Colder

Explanation:

Odyssey ware

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If ram is 40kg and travels 200m in 30 sec find his power

WARRIOR [948]

Work done

N×200
mg200
40(10)(200)
400(200)
80000J

Power

Work done/Time
80000/30
8000/3
2668W

8 0

2 years ago

Read 2 more answers

Whats the Independent and Dependent Variables

astra-53 [7]

The independent variable is the type of fuel used and the dependent variable is the speed of the race car. The independent variable could be changed through the experimental process to see its relation with the dependent variable<span>. The dependent variable is the result of the independent variable changes.</span>

5 0

3 years ago

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

a.

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

b.

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago

Vector A has y-component Ay= +15.0 m . A makes an angle of 32.0 counterclockwise from the +y-axis. What is the x component of A?

Gemiola [76]

Answer:

x-component=-9.3 m

Magnitude of A=17.7m

Explanation:

We are given that

$A_y=+15 m$

$\theta=32^{\circ}$

We have to find the x-component of A and magnitude of A.

According to question

$A_y=\mid A\mid cos\theta$

Substitute the values then we get

$15=\mid A\mid cos32$

$\mid A\mid =\frac{15}{cos32}=\frac{15}{0.848}$

$\mid A\mid=17.7$ m

$tan\theta=\frac{perpendicular\;side}{Base}$

$tan32=\frac{A_x}{A_y}=\frac{A_x}{15}$

$0.62\times 15=A_x$

$A_x=9.3$

The value of x-component of A is negative because the vector A lie in second quadrant.

Hence, the x- component of A=-9.3 m

6 0

3 years ago

Read 2 more answers

What is the correct answer?

pentagon [3]

Answer:

B) x^2+6x+8

Explanation:

x-4 | x^3+2x^2-16x-32

- x^3-4x^2 <-- (x-4)(x^2)

_________________

6x^2-16x-32

- 6x^2-24x <-- (x-4)(6x)

_________________

8x-32

- 8x-32 <- (x-4)(8)

___________________________

0 | x^2+6x+8

This means the answer is B) x^2+6x+8

3 0

2 years ago