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3 years ago

What is the acceleration of a 30.0-kg go kart if the thrust of its engine is 300.0 N?

Physics

2 answers:

VikaD [51]3 years ago

5 0

Answer : $a=10\ m/s^2$

Explanation :

It is given that,

Mass of the engine, m = 30 kg

Thrust is equivalent to the force acting perpendicularly and it is F = 300 N

According to Newton's second law of motion :

$F = m\times a$

a is the acceleration of the engine.

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

$a=\dfrac{300\ N}{30\ Kg}$

$a=10\ m/s^2$

So, the acceleration of the engine is $10\ m/s^2$ .

Hence, this is the required solution.

sp2606 [1]3 years ago

5 0

Answer:

Acceleration, $a=10\ m/s^2$

Explanation:

It is given that,

Mass of the cart, m = 30 kg

Thurst applied to the engine, F = 300 N

Let a is the acceleration of the cart. According to Newton's second law of motion, force is equal to the product of mass and acceleration. It is given by :

$F=ma$

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

$a=\dfrac{300\ N}{30\ kg}$

$a=10\ m/s^2$

So, the acceleration of the cart is $10\ m/s^2$ . Hence, this is the required solution.

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Mass m moves to the right with speed =v along a frictionless horizontal surface and crashes into an equal mass m initially at re

Amiraneli [1.4K]

After the collision the magnitude of the momentum of the system is Mv

Given:

mass of 1st object = M

speed of 1st object = v

mass of 2nd object = M

speed of 2nd object = 0

To Find:

magnitude of the momentum after collision

Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.

Applying conservation of linear momentum

Mv + M(0) = 2MV

Mv = 2MV

V = v/2

So, after collision momentum is

p = 2MV = 2xMxv/2 = Mv

So, after collision momentum is Mv

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1 year ago

What three things can happen to water when it falls on earth's surface

777dan777 [17]

It can be stored on the land surface as ice and snow...it can seep into the earth and be stored as surface water...it can flow in the surface of lands.

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3 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

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

Rearrange this equation to find an equation for $v$ :

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

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

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2 years ago

If all this energy is added to 50 kg of water (the amount of water in a 165-lb person) at 37 ∘C, what is the final state of the

Reika [66]

Answer:

Vapors

Explanation:

We take into account that all the energy from the lightning has been transformed into steam.

$\Delta U = Q - W\\Q = mC \Delta T\\Q = mL$

We calculate the amount of energy required by water to convert into steam.

$Q_{water} = 50 \times \times 4180 \times (100-37)\\= 1.132 \times 10^7 \ J$

$Q_{change\ to\ steam} = 50 \times 2.256 \times 106 \\= 1.128 \times 10^8 \ J$

$Q_{total} = 1.132 \times 10^7 + 1.128 \times 10^8\\= 1.126 \times 10^8 \ J$

From the lightning we received $10^{10} \ J$ of energy, out of which $1.126 \times 10^8$ has been used to convert the water into steam.

Energy left = $10^{10} - 1.126 \times 10^8 = 9.88 \times 10^9 \ J$

We use this energy to convert steam into vapors.

$Q = \Delta E$

$Q = \Delta E = mc (T_{f} - T{i})\\T_{f} = \frac {\Delta E}{mc} + T_{i}\\ \\T_{f} = 100 + \frac{9.88 \times 10^{10}}{50 \times 1970}\\T_{f} = 100 + 10^8\\T_{f} = 10 ^{ \ 8} \ {^ \circ } C$

With this temperature, we can easily interpret that the vapors will be dissociated in hydrogen and oxygen particles.

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2 years ago

If a population is growing what can you conclude about the ammount of resources available

Elodia [21]

Answer:

Generally speaking, as the human population grows, our consumption of natural resources increases. More humans consume more freshwater, more land, more clothing, etc. ... For example, natural gas plants have become increasingly more efficient, thus humans are able to obtain more energy out of the same amount of gas.

Rapid population growth is detrimental to achieving economic and social progress and to sustainable management of the natural resource base. But there remains a sizeable gap between the private and social interest in fertility reduction, and this gap needs to be narrowed.

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2 years ago