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2 years ago

Please help, it's grade 7 science. :)

Physics

2 answers:

ANTONII [103]2 years ago

5 0

I just checked the other persons work and it’s correct you should use their work I was just checking if it was right and if I needed to correct anything

xeze [42]2 years ago

4 0

Answer:

The air in the soccer ball in cold weather will decrease slightly in size and it becomes flat. The air in the soccer ball in hot weather will seem flat because the low preasure leads to lower bounce in the ball.

The metal door frame in cold weather contracts and the wood contracts more in the winter. The metal door frame in hot weather thermal blowing can occur on the outer surface of the metal door frame. Hopefully that is what you were looking for have a good day.

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As you read this in your chair, how fast are you moving relative to the chair? relative to the sun?

Oliga [24]

To answer this question, you need to know the definition of Relative Motion:
The motion is relative when it depends on a reference point or referencial system. If you know the reference point, you can determine the velocity of an object.
If you are sitting on your chair, you are not moving relative to it (Your speed is 0 km/s); but as you know, our planet moves around the Sun (Traslation Movement) with a speed of 30.0 km/s. Therefore, you are moving 30.0 km/s relative to the sun.

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2 years ago

Michael is doing an experiment. He wants to drop a bowling ball and a stuffed bear out the window of a tall building. Under what

Anon25 [30]

Answer:

If there was no air resistance

Explanation:

We know that free fall is a unique motion in which gravity only works on one object. Objects that are said to be free-falling do not experience a significant force of air resistance; They come under the sole effect of gravity. Under such conditions, all objects fall under the same acceleration, regardless of their mass.

6 0

2 years ago

A radioactive substance decays according to the formula w = 20e¡kt grams where t is the time in hours. a find k given that after

blagie [28]

W=20 e(-kt)
A. Rearranging gives k= -(ln(w/20)/t
Substituting w= 10 and solving gives k=0.014
B. Using W=20e(-kt). After 0 hours, W=20. After 24 hours, W=14.29g. After 1 week (24x7=168h) W=1.9g
C. Rearranging gives t=-(ln(10/20)/k. Substituting w=1 and solving gives t=214 hours.
D. Differentiating gives dW/ dt = -20ke(-kt). Solving for t=100 gives dW/dt = 0.07g/h. Solving for t=1000 gives 0.0000002g/h
E. dW/dt = -20ke(-kt). But W=20e(-kt) so dW/dt = -kW

5 0

2 years ago

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between m₁ and m₂ has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between m₁ and m₃ has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where x is measured in m.

The mass m₁ is attracted to m₂ in one direction, and attracted to m₃ in the opposite direction such that m₁ in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for x :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

5 0

2 years ago

The graph at the right shows the force needed to pull a bow back as the string is pulled further and further.

Sindrei [870]

A. 9 J

In a force-distance graph, the work done is equal to the area under the curve in the graph.

In this case, we need to extrapolate the value of the force when the distance is x=30 cm. We can easily do that by noticing that there is a direct proportionality between the force and the distance:

$F=kx$

where k is the slope of the line. We can find k, for instance chosing the point at x=5 cm and F=10 N:

$k=\frac{F}{x}=\frac{10 N}{5 cm}=2 N/cm$

And now we can calculate the work by calculating the area under the curve until x=30 cm, F=60 N:

$W=\frac{1}{2} (height) (base)= \frac{1}{2}(60 N)(0.30 m)=9 J$

B. 24.5 m/s

The mass of the arrow is m=30 g=0.03 kg. The kinetic energy of the arrow when it is released is equal to the work done by pulling back the bow for 30 cm:

$W=K=\frac{1}{2}mv^2$

where m is the mass of the arrow and v is its speed. By re-arranging the formula and using W=9 J, we find the speed:

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2\cdot 9J}{0.03 kg}}=24.5 m/s$

8 0

3 years ago