Which elements are halogens check all that apply

What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?

Assoli18 [71]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

m is the mass

f is the force

From the question we have

$a = \frac{70}{7} = 10 \\$

We have the final answer as

Hope this helps you

4 0

2 years ago

A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D

RideAnS [48]

Answer:

Explanation:

Initial kinetic energy of M = 1/2 M vi²

let final velocity be vf

v² = u² + 2a s

vf² = vi² + 2 (F / M) x D

Kinetic energy

= 1/2 Mvf²

= 1/2 M ( vi² + 2 (F / M) x D

1/2 M vi² + FD

Ratio with initial value

1/2 M vi² + FD) / 1/2 M vi²

RK = 1 + FD / 2 M vi²

4 0

3 years ago

. Write the following in the scientific notation of 0.0000002400m

aleksandr82 [10.1K]

Explanation:

<h2>Steps :</h2>

Move decimal from left to right =0 000000240.0
Then count the numbers before decimal and write it like this =240.0x10 power-9
That's all

hope it help

7 0

2 years ago

The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion.

lesya [120]

Answer:

(a) m = 1.6 x 10²¹ kg

(b) K.E = 2.536 x 10¹¹ J

(c) v = 7.12 x 10⁵ m/s

Explanation:

(a)

First we find the volume of the continent:

V = L*W*H

where,

V = Volume of Slab = ?

L = Length of Slab = 4450 km = 4.45 x 10⁶ m

W = Width of Slab = 4450 km = 4.45 x 10⁶ m

H = Height of Slab = 31 km = 3.1 x 10⁴ m

Therefore,

V = (4.45 x 10⁶ m)(4.45 x 10⁶ m)(3.1 x 10⁴ m)

V = 6.138 x 10¹⁷ m³

Now, we find the mass:

m = density*V

m = (2620 kg/m³)(6.138 x 10¹⁷ m³)

m = 1.6 x 10²¹ kg

(b)

The kinetic energy will be:

K.E = (1/2)mv²

where,

v = speed = (1 cm/year)(0.01 m/1 cm)(1 year/365 days)(1 day/24 h)(1 h/3600 s)

v = 3.17 x 10⁻¹⁰ m/s

Therefore,

K.E = (1/2)(1.6 x 10²¹ kg)(3.17 x 10⁻¹⁰ m/s)²

K.E = 2.536 x 10¹¹ J

(c)

For the same kinetic energy but mass = 77 kg:

K.E = (1/2)mv²

2.536 x 10¹¹ J = (1/2)(77 kg)v²

v = √(2)(2.536 x 10¹¹ J)

v = 7.12 x 10⁵ m/s

7 0

2 years ago

A rock is launched at angle theta=53.2∘ above the horizontal from an altitude of ℎ=182 km with an initial speed ????0=1.61 km/s.

Mariulka [41]

Answer:

The rock's final speed at the required altitude will be 42.24 m/s.

Explanation:

Let's start by finding the initial vertical speed.

Vertical Speed = 1.61 * Sin (53.2°)

Vertical Speed = 0.8 m/s

We want to know the speed of the rock when it is at an altitude of 91 km.

The total displacement of the rock from its starting position will thus be equal to -91 km

We can use this in the following equation:

$s=u*t+\frac{1}{2} (a*t^2)$

$-91=0.8*t+\frac{1}{2} (-9.8*t^2)$

t = 4.3918 seconds

Thus it takes 4.3918 seconds to reach the required altitude. We can now find the speed as follows:

$V=U+at$

$V=0.8+(-9.8)*(4.3918)$

$V = -42.24$

Thus the rock's final speed at the required altitude will be 42.24 m/s.

8 0

3 years ago

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