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2 years ago

Density of water is 1000 kg/m^3. What will be the volume of 35000 kg water?

Physics

1 answer:

d1i1m1o1n [39]2 years ago

8 0

Volume = mass/density
Volume = 35000/1000
Volume = 35m^3

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If a scientist unknowingly breaks the law, he is guilty of _____.

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Answer:

noncompliance

Explanation:

If a scientist unknowingly breaks the law, he is guilty of <u>noncompliance</u>

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3 years ago

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2, High mass stars.

Add-on:

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2 years ago

What is the relationship between Avogadro's number and a mole?

Shtirlitz [24]

Answer:

A. There are 6.02 x 1023 items in a mole, which equals Avogadro's

number

Explanation:

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2 years ago

Two constant forces act on an object of mass m = 5.60 kg object moving in the xy plane as shown in the figure below. Force F1 is

Gnoma [55]

Answer:

Part a)

$F_1 = 22.5\hat i + 15.7 \hat j$

$F_2 = -35.5 \hat i + 20.5 \hat j$

Part b)

$F = -10 \hat i + 36.2 \hat j$

Part c)

$a = -1.78 \hat i + 6.46 \hat j$

Part d)

$v_f = -0.64 \hat i + 21.93\hat j$

Part e)

$r_f = 6.09\hat i + 36.72 \hat j$

Part f)

$KE = 1347.7 J$

Part g)

$KE = 1348 J$

Explanation:

Part a)

first force is given as

$F_1 = 27.5 N$ at 35 degree

$F_1 = 27.5 cos35 \hat i + 27.5 sin35 \hat j$

$F_1 = 22.5\hat i + 15.7 \hat j$

Second force is given as

$F_2 = 41 N$ at 150 degree

$F_2 = 41 cos150 \hat i + 41 sin150\hat j$

$F_2 = -35.5 \hat i + 20.5 \hat j$

Part b)

Total force is given as

$F = F_1 + F_2$

$F = (22.5 - 35.5)\hat i + (15.7 + 20.5)\hat j$

$F = -10 \hat i + 36.2 \hat j$

Part c)

as we know

F = ma so object acceleration is given as

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

$a = \frac{-10}{5.60} \hat i + \frac{36.2}{5.60} \hat j$

$a = -1.78 \hat i + 6.46 \hat j$

Part d)

By kinematics we know that

$v_f = v_i + at$

$v_f = (4.70 \hat i + 2.55 \hat j) + (-1.78\hat i + 6.46\hat j)(3)$

$v_f = -0.64 \hat i + 21.93\hat j$

Part e)

As we know that final position is given as

$r_f = v_i t + \frac{1}{2}at^2$

$r_f = (4.70 \hat i + 2.55 \hat j)(3) + \frac{1}{2}(-1.78 \hat i + 6.46 \hat j)(3^2)$

$r_f = 6.09\hat i + 36.72 \hat j$

Part f)

final kinetic energy is given as

$KE = \frac{1]{2}mv_f^2$

$KE = \frac{1}{2}(5.60)(0.64^2 + 21.93^2)$

$KE = 1347.7 J$

Part g)

final kinetic energy is given as

$KE = KE_i + F.r$

$KE = \frac{1}{2}(5.60)(4.70^2 + 2.55^2) + (-10 \hat i + 36.2 \hat j).(6.09\hat i + 36.72\hat j)$

$KE =80 + 1268.4$

$KE = 1348 J$

4 0

3 years ago

A football is thrown at an angle of 44◦ above the horizontal. Assume the ball is throw and received at the same height. To throw

aleksley [76]

Answer:

The initial speed of the ball is 23.3 m/s.

Explanation:

Given that,

Angle = 44°

Range = 55.7 m

We need to calculate the initial speed of the ball

Using formula of range

$R=\dfrac{u^2\sin2\theta}{g}$

Where. u = initial velocity

r = range

g = acceleration due to gravity

Put the value into the formula

$55.7=\dfrac{u^2\sin2\times44}{9.8}$

$u^2=\dfrac{55.7\times9.8}{\sin88}$

$u=\sqrt{\dfrac{55.7\times9.8}{\sin88}}$

$u=23.3\ m/s$

Hence, The initial speed of the ball is 23.3 m/s.

4 0

2 years ago