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3 years ago

The acceleration due to gravity on Earth is 9.80 m/s 2 . An African elephant can have a mass up to 6,050 kg. What is its weight

Physics

2 answers:

kirza4 [7]3 years ago

8 0

Answer:

59,290 N

Explanation:

The weight of an object is given by:

$W=mg$

where

m is the mass of the object

g is the acceleration due to gravity

In this problem:

m = 6050 kg

g = 9.80 m/s^2

Substituting into the equation, we find:

$W=(6050 kg)(9.80 m/s^2)=59,290 N$

Vlada [557]3 years ago

4 0

The answer to this is 59,290 N

Hope this helped :)

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A ball is thrown up into the air with 100 j of kinetic energy, which is transformed to gravitational potential energy at the top

jenyasd209 [6]

The kinetic energy when it returns to its original height is 100 J

Solution:

The ball is thrown up with a Kinetic Energy K. E. = 0.5×m×v² = 100 J

Therefore the final height is given by

Where:

u = final velocity = 0

v = initial velocity

s = final height

Therefore v² = 2·g·s = 19.62·s

P.E = Potential Energy = m·g·s

Since v² = 2·g·s

Substituting the value of v² in the kinetic energy formula, we obtain

K. E. = 0.5×m×2·g·s = m·g·s = P.E. = 100 J

When the ball returns to the original height, we have

v² = u² + 2·g·s

Since u = 0 = initial velocity in this case we have

v² = 2·g·s and the Kinetic energy = 0.5·m·v²

Since m and s are the same then 0.5·m·v² = 100 J.

As the height of the ball increases the kinetic energy of the ball is converted into gravitational potential energy. This means that the kinetic energy of the bullet is reduced. When the ball reaches its maximum height, it momentarily comes to rest and the ball's kinetic energy is zero. When the ball hits the ground, its potential energy is converted to kinetic energy.

Learn more about Kinetic energy here:-brainly.com/question/25959744

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6 0

1 year ago

Given: Saturated air changes temperature by 0.5°C/100 m. The air is completely saturated at the dew point. The dew point has bee

erma4kov [3.2K]

Answer:

Explanation:

Given

saturated air temperature by $0.5^{\circ}C/100 m$

Dew point temperature is given by $t=2^{\circ}C$

Dew point is defined as the temperature after which air no longer to uphold the water vapor fuse with it and some water vapor may condense to a liquid.

air continues to rise for 1400 m

i.e. change in temperature would be $\Delta t =\frac{0.5}{100}\times 1400=7^{\circ}C$

Final temperature $t_f$

$t_f+\Delta t=t$

$t_f=2-7=-5^{\circ}C$

3 0

3 years ago

A 0.1 kg toy contains a compressed spring. when the spring is released the toy fly 0.45 m upwards from ground level before falli

trapecia [35]

The speed of the toy when it hits the ground is 2.97 m/s.

The given parameters;

mass of the toy, m = 0.1 kg

the maximum height reached by the, h = 0.45 m

The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.

P.E = K.E

$mgh_{max} = \frac{1}{2} mv_{max}^2\\\\gh_{max} = \frac{1}{2} v_{max}^2\\\\v_{max}^2= 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}}$

Substitute the given values and solve the speed;

$v_{max} = \sqrt{2\times 9.8 \times 0.45} \\\\v_{max} = 2.97 \ m/s$

Thus, the speed of the toy when it hits the ground is 2.97 m/s.

Learn more here: brainly.com/question/7562874

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3 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .