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alexandr402 [8]

3 years ago

11

An object located near the surface of Earth has a weight of a 245 N

1 answer:

gayaneshka [121]3 years ago

7 0

Answer:

ثر أنواع التربة خصوبة التربحمراء .

ج- السوداء

Explanation:

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A 40.0-$kg$ body is moving in the direction of the positive x axis with a speed of 238 $m/s$ when, owing to an internal explosio

Julli [10]

Answer:

$v_3 = 400\ m/s$

Explanation:

given,

mass of the body = 40 Kg

speed in x-axis = 238 m/s

mass break into three part

m₁ = 7 kg

v₁ = 356 m/s (along the positive y axis)

m₂ = 4.5 kg

v₂ = 357 m/s(along the negative x axis)

m₃ = 40 - (7 + 4.5) = 28.5 Kg

v₃ = ?

using conservation of momentum

MV = m₁v₁ + m₂v₂ + m₃v₃

$(40)(238) \hat{i} = (7)(356) \hat{j} - (4.5)(357) \hat {i} + 28.5 v_3$

$(9520) \hat{i} = 2492 \hat{j} - 1606.5\hat {i} + 28.5 v_3$

$11126.5 \hat{i} - 2492 \hat{j} = 28.5 v_3$

$v_3 =390.40 \hat{i} - 87.44 \hat{j}$

$v_3 = \sqrt{390.40^2 + 87.44^2}$

$v_3 = 400\ m/s$

4 0

3 years ago

Consider the two-body situation at the right. A 3.50x103-kg crate (m1) rests on an inclined plane and is connected by a cable to

4vir4ik [10]

Answer: $a=2.42 m/s^2$

Explanation:

Given

mass $m_1=3.50\times 10^{3} kg$

$m_2=1.00\times 10^{3} kg$

$\theta$ (inclination)= $30^{\circ}$

$\mu =0.21$

Let T be the tension in the rope

From Diagram

$m_1g\sin \theta -T-f_r=m_1a$ -----------------1

where $f_r=friction\ force$

$f_r=\mu m_g\cos \theta$

For block $m_2$

$T=m_2a$ -----------2

From 1 & 2

$m_1g\sin \theta -m_2a-\mu m_1g\cos \theta =m_1a$

$m_1g(\sin \theta -\mu \cos \theta )=(m_1+m_2)a$

$\frac{9.8\times 3.5\times 10^3}{4.5\times 10^3}(0.5-\0.21\cos 30)=4.5a$

$a=2.42 m/s^2$

4 0

4 years ago

What is the magnitude of the kinetic frictional force

Effectus [21]

The magnitude of the kinetic friction force, ƒk, on an object is. Where μk is called the kinetic friction coefficient and |FN| is the magnitude of the normal force of the surface on the sliding object. The kinetic friction coefficient is entirely determined by the materials of the sliding surfaces. hope it helps

4 0

3 years ago

Select the correct answer.

TEA [102]

OA the speed will increase four times

4 0

3 years ago

Read 2 more answers

A uniform solid disk and a uniform ring are place side by side at the top of a rough incline of height h.

nadezda [96]

Explanation:

velocity of disc $=\sqrt((gh)/0.75)$

lets call (h) 1 m to make it simple.

= 3.614 m/s

$\sqrt((4/3) x 1 x 9.8) = 3.614$ m/s pointing towards this:

$4×V_d=\sqrt(4/3hg)$

$V_h=\sqrt(hg)$

velocity of hoop= $\sqrt(gh)$

lets call (h) 1m to make it simple again.

$\sqrt(9.8 x 1) = 3.13$ m/s

$\sqrt(gh) = sqrt(hg)so [tex]4×V_d= \sqrt(4/3hg)V_h=\sqrt(hg)$

The disc is the fastest.

While i'm on this subject i'll show you this:

Solid ball $=0.7v^2= gh$

solid disc $= 0.75v^2 = gh$

hoop $=v^2=gh$

The above is simplified from linear KE + rotational KE, the radius or mass makes no difference to the above formula.

The solid ball will be the faster of the 3, like above i'll show you.

solid ball: velocity $=\sqrt((gh)/0.7)$

let (h) be 1m again to compare.

$\sqrt((9.8 x 1)/0.7) = 3.741$ m/s

solid disk speed $=\sqrt((gh)/0.75)$

uniform hoop speed $=\sqrt(gh)$

solid sphere speed $=\sqrt((gh)/0.7)$

8 0

3 years ago