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3 years ago

An object, initially at rest, is subject to an acceleration of 34 m/s2. how long will it take for that object to travel 3400 m?

Physics

1 answer:

Montano1993 [528]3 years ago

4 0

15 seconds i may be wrong though

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I need help ASAP please :)

rusak2 [61]

Density offers a convenient means of obtaining the mass of a body from its volume or vice versa; the mass is equal to the volume multiplied by the density (M = Vd), while the volume is equal to the mass divided by the density (V = M/d).

M = V d

M = 1.4 * 2 = 2.8 kg

7 0

2 years ago

The roller coaster is designed according to safety regulations that prohibit the speed of the car from exceeding 20 m/s. find th

hram777 [196]

maximum allowed value of the speed in roller coaster is given as

$v = 20 m/s$

now from kinematics we can say

$v^2 - v_i^2 = 2 a s$

here initial speed will be

$v_i = 0$

acceleration is due to gravity

$a = 9.8 m/s^2$

now we can use this to find the height

$20^2 - 0^2 = 2 * 9.8* h$

$400 = 19.6 *h$

$h = 20.4 m$

so maximum allowed height will be 20.4 m

4 0

3 years ago

What is the volume of a bar of soap that is 9 cm long, 5 cm wide, and 2 cm high?

GalinKa [24]

Answer:

90cm

Explanation:

2x5=10

10x9=90

7 0

3 years ago

Read 2 more answers

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$