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3 years ago

Please help i’m running out of time to answer

Physics

1 answer:

kipiarov [429]3 years ago

4 0

Answer:

5,970 N

Explanation:

m = 597 kg

a = 10 m/s^2

Plug those values into the following equation:

F = ma

F = (597 kg)(10 m/s^2)

F = 5,970 N

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An object is released from rest and falls in free fall motion. The speed v of the object after it has fallen a distance y is giv

konstantin123 [22]

Answer:

8.91 %

Explanation:

Since v² = 2gy

By the relative error formula,

2Δv/v = Δg/g + Δy/y multiplying by 100%, we have

2Δv/v × 100% = Δg/g × 100 % + Δy/y × 100%

2(Δv/v × 100%) = Δg/g × 100 % + Δy/y × 100%

Δg/g × 100 % = 2(Δv/v × 100%) - Δy/y × 100%

Since Δv/v × 100% = 3.69 % and Δy/y × 100% = 5 %

Since we have a difference for the percentage error in g, we square the percentage errors and add them together. So,

[Δg/g × 100 %]² = [2(Δv/v × 100%)]² + [Δy/y × 100%]²

[Δg/g × 100 %]² = [2(3.69)]² + [5%]²

[Δg/g × 100 %]² = [4)(3.69 %)² + [5%]²

[Δg/g × 100 %]² = 54.4644 %² + 25%²

[Δg/g × 100 %]² = 79.4644 %²

taking square-root of both sides, we have

[Δg/g × 100 %] = 8.91 %

So, the percent uncertainty in the calculated value of g is 8.91 %

6 0

3 years ago

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

aliina [53]

Answer:

9.43 m/s

Explanation:

First of all, we calculate the final kinetic energy of the car.

According to the work-energy theorem, the work done on the car is equal to its change in kinetic energy:

$W=K_f - K_i$

where

W = -36.733 J is the work done on the car (negative because the car is slowing down, so the work is done in the direction opposite to the motion of the car)

$K_f$ is the final kinetic energy

$K_i = 66,120 J$ is the initial kinetic energy

Solving,

$K_f = K_i + W = 66,120 + (-36,733)=29,387 J$

Now we can find the final speed of the car by using the formula for kinetic energy

$K_f = \frac{1}{2}mv^2$

where

m = 661 kg is the mass of the car

v is its final speed

Solving for v, we find

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.43 m/s$

3 0

3 years ago

The tundra is located _______.

Leona [35]

Answer:

d: both north and south

Explanation:

8 0

3 years ago

Read 2 more answers

E asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.0 earth years. part a wha

anzhelika [568]

To solve the problem, use Kepler's 3rd law :

T² = 4π²r³ / GM

Solved for r :

r = [GMT² / 4π²]⅓

but first covert 6.00 years to seconds :

6.00years = 6.00years(365days/year)(24.0hours/day)(6...
= 1.89 x 10^8s

The radius of the orbit then is :

r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓
= 6.23 x 10^11m

6 0

3 years ago

Compare the speed of light in water to the speed of sound in water

kakasveta [241]

Answer:As a rule sound travels slowest through gases,faster through liquids,and fastest through solids.The speed of light as it travels through air and space is much faster than that of sound; it travels at 300 million meters per second or 273,400 miles per hour.Speed of light in water = 226 million m/s or 205,600 mph.

Explanation:dont have one

6 0

3 years ago