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3 years ago

Hellllppppppp please

Physics

1 answer:

WITCHER [35]3 years ago

3 0

I hope that helps! Let me know if you need any more help!

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A small toy car travels 2.0 meters in 120 seconds. what is the speed of the car?

mamaluj [8]

Answer: $\dfrac{1}{60} ms^{-1}$

Explanation:

Speed of an object is defined as the ratio of the distance covered by the object to the time taken to cover that distance.

Let $s$ be the speed of the object.

Let $D$ be the distance travelled by the object.

Let $t$ be the time taken by the object.

So, $s=\frac{D}{t}$

s= $\frac{2}{120}=\frac{1}{60}ms^{-1}$

So,the speed of the car is $\frac{1}{60}ms^{-1}$

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3 years ago

Calculate the density of a solid cube that

tia_tia [17]

Ok so if each side is 4.53 cm, we can multiply 4.53 x 4.53 x 4.53 to get the volume (since v= l x w x h). Density equals mass/volume, so

519 g/4.53 cm
114.57 g/cm^3 (since none of the units cancel)

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3 years ago

Which of the following BEST describes what a thermometer measures

asambeis [7]

A theromometer is the increase or decrease of earths atmospheric temperture, thats how you would measure the temperture of the air around you.

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3 years ago

Gas hydrates are used _____.

Nonamiya [84]

Hope this helps mark brainest plz

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3 years ago

Read 2 more answers

Daring Darless wishes to cross the Grand Canyon of the Snake River by being shot from a cannon. She wishes to be launched at 65

guajiro [1.7K]

Answer:

She must be launched with minimum speed of <u>57.67 m/s</u> to clear the 520 m gap.

Step-by-step explanation:

Given:

The angle of projection of the projectile is, $\theta =65$ °

Range of the projectile is, $R=520$ m.

Acceleration due to gravity, $g=9.8\ m/s^2$

The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.

The range of projectile is given as:

$R=\frac{v_{0}^2\sin2\theta}{g}$

Plug in all the given values and solve for minimum speed, $v_0$ .

$520=\frac{v_{0}^2\sin(2(65))}{9.8}\\520\times 9.8=v_{0}^2\sin(130)\\5096=1.532v_{0}^2\\v_0^2=\frac{5096}{1.532}\\v_0^2=3326.371\\v_0=\sqrt{3326.371}=57.67\textrm{ m/s}$

Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.

3 0

3 years ago