In Figure 4.27 four particles form a square of edge length

Which statement is true about the theory of plate tectonics and the theory of continental drift?:

Rufina [12.5K]

The answer is A I had this question. Before

5 0

2 years ago

A ball is swung in a horizontal circle at a constant speed. Each circle takes 0.85 seconds to complete and the rope is 0.40 m lo

BaLLatris [955]

Answer:

The centripetal acceleration will be "21.785 m/s²".

Explanation:

The given values are:

Time,

t = 0.85 seconds

Length of rope,

r = 0.40 m

Mass of ball,

m = 0.80 kg

As we know,

⇒ $w=\frac{2 \pi}{t}$

On substituting the values, we get

⇒ $=\frac{2\times 3.14}{0.85}$

⇒ $= \frac{6.28}{0.85}$

⇒ $=7.38 \ rad/s^2$

The centripetal acceleration will be:

⇒ $a=r\times w^2$

⇒ $=0.40\times (7.38)^2$

⇒ $=0.40\times 54.46$

⇒ $=21.785 \ m/s^2$

7 0

2 years ago

. A 1.00 kg rock is thrown up into the air from ground level at a speed of 8.00 m/s. The ball travels up

Liula [17]

Answer:

p = -8 kg-m/s

Explanation:

Given that,

Initial speed of the rock, u = 8 m/s

Mass of the rock, m = 1 kg

The ball travels up to a maximum height, then returns to the ground.

We need to find the rock's momentum as it strikes the ground. Let v be the final speed of the rock. Its final speed is as same as initial speed i.e. 8 m/s but in negative direction. So

p = mv

p = 1 kg × (-8 m/s)

= -8 kg-m/s

So, the rock's momentum as it strikes the ground is (-8 kg-m/s).

5 0

3 years ago

A piece of metal is 4 cm by 10 cm by 2 cm. Find its mass if it has a density of 2 g/cm'.

harina [27]

Answer:

m=ρV

V=LxWxD

V=4x10x2=80

m=2*80=160 grams

Explanation:

Mass is equal to density multiply by volume of object. Volume of rectangle pice can be calculated by multiplying all sides.

3 0

3 years ago

Traumatic brain injury such as a concussion results when the head undergoes a very large acceleration. Generally an acceleration

eimsori [14]

The complete text of the problem is:

"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "

Solution:

1) Acceleration: $-2336 m/s^2$ on the hardwood floor, $-382 m/s^2$ on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

$v^2 - u^2 = 2ad$

where

v is the final speed

u = 0 is the initial speed (the child starts from rest)

a = g = 9.8 m/s^2 is the acceleration of gravity

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

$v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s$

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

$d = 1.8 mm = 0.0018 m$

So we can find the acceleration by using again the equation

$v'^2 - v^2 = 2ad$

Solving for a,

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2$

For the carpeted floor instead,

$d=1.1 cm = 0.011 m$

therefore the acceleration is

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2$

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

$a=\frac{v'-v}{t}$

where

v' = 0

v = 2.9 m/s

For the hardwood floor,

$a=-2336 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-2336}=0.00124 s = 1.24 ms$

For the carpeted floor,

$a=-382 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-382}=0.00759 s = 7.59 ms$

We can now comment the results using the initial statement of the problem:

"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

8 0

3 years ago