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2 years ago

A cheetah ran 300 feet in 2.92 seconds. What was the cheetah’s average speed in miles per hour?

1 answer:

6 0

We will first convert feet to miles:
1 foot = 0.00018939 miles
300 foot = 0.056817 miles
Than seconds to hours:
2.92 seconds = 0.04867 minutes = 0.0008111 hours
Average speed :
v = d / t
v = 0.056817 miles / 0.008111 h
v = 70.05 miles per hour

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Asteroids are between 1000 km and less than 10 m in diameter. What is the diameter of most asteroids?

e-lub [12.9K]

This question needs research to be answered. From the given information alone it can't be answered without making wild assumptions.

Ideally, you need to take a look at a distribution (or a histogram) of asteroid diameters, identify the "mode" of such a distribution, and find the corresponding diameter. That value will be the answer.

I am attaching one such histogram on asteroid diameters from the IRAS asteroid catalog I could find online. (In order to get a single histogram, you need to add the individual curves in the figure first). Eyeballing this sample, I'd say the mode is somewhere around 10km, so the answer would be: the diameter of most asteroid from the IRAS asteroid catalog is about 10km.

7 0

2 years ago

Which has greater buoyant force, light corn syrup or the vegetable oil?

natta225 [31]

It is most likely vegetable oil

6 0

3 years ago

Projectiles that strike objects are good examples of inelastic collisions. A 0.1 kg nail driven by a gas powered nail driver col

Ratling [72]

In an inelastic collision, only momentum is conserved, while energy is not conserved.

1) Velocity of the nail and the block after the collision
This can be found by using the total momentum after the collisions:
$p_f=(m+M)v_f=4.8 kg m/s$
where
m=0.1 kg is the mass of the nail
M=10 kg is the mass of the block of wood
Rearranging the formula, we find $v_f$ , the velocity of the nail and the block after the collision:
$v_f= \frac{p_f}{m+M}= \frac{4.8 kg m/s}{0.1 kg+10 kg}= 0.48 m/s$

2) The velocity of the nail before the collision can be found by using the conservation of momentum. In fact, the total momentum before the collision is given only by the nail (since the block is at rest), and it must be equal to the total momentum after the collision:
$p_i = mv_i = p_f$
Rearranging the formula, we can find $v_i$ , the velocity of the nail before the collision:
$v_i = \frac{p_f}{m}= \frac{4.8 kg m/s}{0.1 kg}=48 m/s$

6 0

2 years ago

Read 2 more answers

An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.

Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

$The \ weight, W =G\dfrac{M \times m}{R^{2}} = 800 \ N$

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

$W_1 =G\dfrac{\dfrac{M}{2} \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2} \times m \times 4}{ R ^{2}} = 2 \times G \times \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N$

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0

2 years ago

A particle on a spring moves in simple harmonic motion along the x axis between turning points at x1 = 95 cm and x2 = 135 cm. (i

uranmaximum [27]

Answer:

(i) $x = 115\,cm$ , (ii) $x = 95\,cm$ , (iii) $x = 95\,cm$

Explanation:

(i) $x_{1}$ and $x_{2}$ represent the points where particle has a velocity of zero and spring reach maximum deformation, Given the absence of non-conservative force and by the Principle of Energy Conservation, the position where particle is at maximum speed is average of both extreme positions:

$x = 115\,cm$

(ii) Maximum accelerations is reached at $x_{1}$ and $x_{2}$ .

$x = 95\,cm$

(iii) Greatest net forces exerted on the particle are reached at $x_{1}$ and $x_{2}$ .

$x = 95\,cm$

8 0

2 years ago