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2 years ago

Can anyone help me in this question?

Physics

1 answer:

Anna71 [15]2 years ago

3 0

Answer:

I guess the answer is charging by friction

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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

2 years ago

Highest common factor of 12r and 10

Scilla [17]

I think the number 2, not sure

3 0

1 year ago

A Hooke's law spring is compressed 12.0 cm from equilibrium, and the potential energy stored is 72.0 J. What compression (as mea

Anuta_ua [19.1K]

Answer:14.14 cm

Explanation:

Given

Spring Compression $x=12 cm$

Potential energy Stored in spring $=72 J$

Suppose k is the spring constant of spring

Potential Energy of spring is given by $=\frac{kx^2}{2}$

$\frac{k(0.12)^2}{2}=72$

$k(0.12)^2=144$

$k=10,000 N/m$

$k=10 kN/m$

for 100 J energy

$\frac{k(x_0)^2}{2}=100$

$10\times 10^3\cdot (x_0)^2=200$

$(x_0)^2=2\times 10^{-2}$

$x_0=0.1414$

$x_0=14.14 cm$

6 0

2 years ago

In 1970, a rocket powered car called Blue Flame achieved a maximum speed of 1.00(10 km/h (278m/s).Suppose the magnitude of the c

sammy [17]

Answer:

Distance traveled during this acceleration will be 6950 m

Explanation:

Wear have given maximum speed tat will be equal to final speed of the car v = 278 m/sec

Constant acceleration $a=5.56m/sec^2$

As the car starts initially starts from rest so initial velocity of the car u = 0 m/sec

From third equation of motion $v^2=u^2+2as$

Putting all values in equation

$278^2=0^2+2\times 5.56\times s$

s = 6950 m

So distance traveled during this acceleration will be 6950 m

3 0

2 years ago

A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on

n200080 [17]

Answer:

6858.5712 m/s

Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

V = 6858.5712 m/s

6 0

2 years ago