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1 year ago

10

PLEASE HELP-I LITERALLY NEED THIS RIGHT NOW!!! I'LL GIVE BRAINLIEST AND 5 STARS!!!!!!!!!!!!!!!!

2 answers:

SOVA2 [1]1 year ago

4 0

Answer:

Oceanic crust moved over the hot spot, each recently formed volcano was carried away from the hot spot toward the northwest, cutting off its source of lava. Meanwhile, a new island was forming so that over time a chain of islands was produced extending away from the hot spot.

valkas [14]1 year ago

3 0

Answer: oceanic crust moved over the hot spot, each recently formed volcano was carried away from the hot spot toward the northwest, cutting off its source of lava. Meanwhile, a new island was forming so that over time a chain of islands was produced extending away from the hot spot.ANSWER HERE: This mean that volcanos are producing these new islands and it is creating a chain of different islands all over Hawaii.

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Will Give Brainlist <br> Check pic for question

Harrizon [31]

Ohms law = v= Ir

V= 0.02 x 4000 = 80v

5 0

2 years ago

Read 2 more answers

Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v

bazaltina [42]

Answer:

1. $t_2 = 2t_1$

2. $t_2 = t_1\sqrt{2}$

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

$a_1 = 2a_2$

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

$t = v / a_1$

Now that acceleration is halved:

$t = \frac{v}{2a_2}$

$\frac{v}{a_2} = 2t$

You would need to push for twice amount of time $t_2 = 2t_1$

2. The distance traveled by the puck is as the following equation:

$d = at^2$

So if the acceleration is halved while maintaining the same d:

$\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}$

As $d_1 = d_2$ , then $d_1/d_2 = 1$ . Also $a_1 = 2a_2$

$1 = \frac{2a_2t_1^2}{a_2t_2^2}$

$t_2^2 = 2t_1^2$

$t_2 = t_1\sqrt{2}\approx 1.14t_1$

So t increased by 1.14

7 0

3 years ago

What is the wavelength of radar waves with the frequency of 6.0 x 10 ^10 Hz?

Serjik [45]

Yes yes multiply hurry up

8 0

2 years ago

Density=2g/mL and volume=20mL what is mass

leva [86]

Answer:

40g

Explanation:

Mass = density x volume

= 2 x 20

= 40g

8 0

3 years ago

Read 2 more answers

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0

2 years ago