All categories

3 years ago

Can someone please answer this, ill give you brainliest Would be very appreciated.

Physics

2 answers:

Vilka [71]3 years ago

6 0

Answer:

Snail shells

Explanation:

Yea surprising answer but like this is from the web -> "New research shows that snail shells retain their carbon and oxygen isotope signatures after being boiled and, through those proxies, likely still preserve an accurate record of past temperature, rainfall, and vegetation."

Sorry if wrong and hope it helps! :)

Deffense [45]3 years ago

3 0

Answer:

leaf margins

Explanation:

Scientists study ice cores and leaf margins to learn about the past climates.

Further explanation :

Leaf margins are an indicator of the temperature in which a tree lives or lived. Paleobotanists use the margins of fossil leaves to infer past temperatures.

You might be interested in

A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta

Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, $\theta = 54^{\circ}$

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, $R = \frac{v^{2}sin2\theta}{g}$

$R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m$

Time of flight, t' = $\frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s$

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

$s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m$

Distance traveled vertically:

$s_{y} = vcos\theta - \frac{1}{2}gt^{2}$

$s_{y} = 12.4sin54^{\circ}\times 1.42 - \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m$

Now,

$s_{x} = \frac{mx + m'x'}{M}$

$10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}$

x' = 3.535 m

Similarly,

$s_{y} = \frac{my + m'y'}{M}$

$4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m$

The location of the other fragment is at (3.535, 1.162)

5 0

4 years ago

HELP ASAP!!!

butalik [34]

Answer:

D. The acceleration after it leaves the hand is 10 m/s/s downwards

Explanation:

Vertical Throw

When an object is thrown upwards, it describes a special type of motion ruled only by gravity.

When the ball is launched, it has its maximum speed upwards. The acceleration of gravity is always the same because it's a constant value near our planet's surface. The object starts to lose speed since the acceleration of gravity is pointed downwards and makes the object stop in the mid-air at its maximum height, where the speed is zero. Then, the object starts to fall and regain speed, this time downwards until it reaches back the launching point at the very same speed it was launched, but in the opposite direction.

The time it takes to reach its maximum height is the same it takes to return to the catching point, 2 seconds later.

With all these concepts in mind, we state that:

D. The acceleration after it leaves the hand is 10 m/s/s downwards

The other options are not correct because:

A. The acceleration is never upwards

B. The acceleration is never 0

C. Both times are equal

8 0

3 years ago

If the retina is 1.7 cm from the lens in the eye, how large is the image on the retina of a person of height 1.5 m standing 9.0

sineoko [7]

Answer:

The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign

Explanation:

u = Object distance = 9 m

v = Image distance = 1.7 cm (as the image is forming on the retina)

$h_u$ = Object height = 1.5 m

Magnification

$m=-\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{0.017}{9}=\frac{h_v}{1.5}\\\Rightarrow h_v=-0.00283\ m$

The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign

4 0

3 years ago

n this experiment, you will measure mass and volume of various sport balls, both regularly shaped ball and irregularly shaped. W

pentagon [3]

Explanation :

Density of an object can determine using mass and volume. We can measure mass and volume of various sports ball, both regular shaped ball and irregular shaped ball.

Density for regularly shaped ball and that of irregularly shaped ball is different.

Suppose that mass of an object is 100g and the volume is $2000\ cm^3$ . So, using this data we can find density of that object as