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Ghella [55]
4 years ago
7

Which of the following is an element?

Physics
1 answer:
oksano4ka [1.4K]4 years ago
7 0
Chlorine pls mark as brianliest
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A man tries to push a 200 kg Car that moves at a acceleration 0.50 m/s2. The man is able to displace the car 10 m. How much work
yawa3891 [41]

The work done by the man pushing the car over the given distance is 1000J.

Given the data in the question;

  • Mass of car; m = 200kg
  • Acceleration of the car; a = 0.5m/s^2
  • Distance covered by the car; d = 10m

Work done; W = \ ?

<h3>Work done</h3>

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

Work\ done = f * d

Where f is force applied and d is distance travelled.

To determine the work done by the man, we first solve for the force applied F.

From Newton's Second Law; Force \ F = m * a

We substitute our given values into the expression

F = m * a \\\\F = 200kg * 0.5m/s^2\\\\F = 100kg.m/s^2

Next we substitute our values into the expression of work done above.

Work \ done = f * d\\\\Work \ done = 100kg.m/s^2 * 10m\\\\Work \ done = 1000kgm^2/s^2\\\\Work \ done = 1000J

Therefore, the work done by the man pushing the car over the given distance is 1000J.

Learn more about work done: brainly.com/question/26115962

7 0
2 years ago
What is a way to make magnets? <br><br> Heating<br> Pressure<br> Freezing
AlladinOne [14]

Answer:

The best easy to make magnrets is by heating it

4 0
3 years ago
Read 2 more answers
the pull of gravity on mars is 3.7m/s^2. if a astronaut on mars lifts a 10 kg rock 1 m off the ground, just to see whats under i
Elanso [62]
Gravitational potential energy can be calculated using the formula:

PE_{grav} =mgh

Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height

On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.

So put in what you know and solve for what you don't know. 
m = 10kg
g = 3.7m/s^2
h = 1m

So we put that in and solve it. 
PE_{grav} =mgh
PE_{grav} =(10kg)(3.7m/s^{2})(1m)
PE_{grav} =37J


7 0
4 years ago
A lead ball is dropped into a lake from a diving board 5.0 meters above the water. After entering the water, it sinks to the bot
xxMikexx [17]

Answer:

28.8 meters

Explanation:

We must first determine at which velocity the ball hits the water. To do so we will:

1) Assume no air resistance.

2) Use the Law of conservation of mechanical energy: E=K+P

Where

E is the mechanical energy (which is constant)

K is the kinetic energy.

P is the potential energy.

With this we have \frac{m}{2} *v^{2}  = m*g*h

Where:

m is the balls's mass <- we will see that it cancels out and as such we don't need to know it.

v is the speed when it hits the water.

g is the gravitational constant (we will assume g=9.8\frac{m}{s^{2} }.

h is the height from which the ball fell.

Because when we initially drop the ball, all its energy is potential (and P = - m*g*h) and when it hits the water, all its energy is kinetic (K=\frac{m}{2} *v^{2}. And all that potential was converted to kinetic energy.

Now, from \frac{m}{2} *v^{2}  = m*g*h we can deduce that v=\sqrt{2*g*h}

Therefore v=9.6\frac{m}{s}

Now, to answer how deep is the lake we just need to multiply that speed by the time it took the ball to reach the bottom.

So D=9.6\frac{m}{s}*3s=28.8m

Which is our answer.

7 0
3 years ago
Read 2 more answers
Sarah is making bread. She measures 0.6 kg of hot water into a bowl. The water needs to be at a temperature of
Mrrafil [7]

Answer:

\Delta m = 0.171\,kg

Explanation:

The process of water adding is described by the First Law of Thermodynamics:

m_{w,o} \cdot h_{w,o} + \Delta m \cdot h_{w} = (m_{w,o} + \Delta m)\cdot h_{w,f}

The amount of additional mass is:

m_{w,o}\cdot h_{w,o} - m_{w,o}\cdot h_{w,f} = \Delta m\cdot (h_{w,f}-h_{w})

\Delta m = \frac{m_{w,o}\cdot(h_{w,o}-h_{w,f})}{h_{w,f}-h_{w}}

Given that water is incompressible, the equation can be further simplified:

\Delta m = m_{w,o}\cdot \frac{c_{p,w}\cdot (T_{w,o}-T_{w,f})}{c_{p,w}\cdot (T_{w,f}-T_{w})}

\Delta m = m_{w,o}\cdot \left(\frac{T_{w,o}-T_{w,f}}{T_{w,f}-T_{w}} \right)

\Delta m = (0.6\,kg)\cdot \left(\frac{40 ^{\circ}C - 42^{\circ}C}{42^{\circ}C-49^{\circ}C} \right)

\Delta m = 0.171\,kg

.

6 0
3 years ago
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