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2 years ago

11

A car, starting from rest, accelerates in a straight-line path at a constant rate of 2.5 m/s2. How far will the car travel in 12

seconds?

1 answer:

Ksenya-84 [330]2 years ago

6 0

Answer:

180 m

Explanation:

i found the formula

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(a) Calculate the buoyant force on a 2.00-L helium balloon.

iragen [17]

Answer:

0.0239364 N

0.0057879 N

Explanation:

$\rho$ = Density of the gas

g = Acceleration due to gravity = 9.81 m/s²

V = Volume

Mass of rubber = 1.5 g

Buoyant force is given by

$F_b=\rho gV\\\Rightarrow F_b=1.22\times 9.81\times 2\times 10^{-3}\\\Rightarrow F_b=0.0239364\ N$

The buoyant force is 0.0239364 N

Net vertical force is given by

$F_n=F_b-W_{He}-W_{r}\\\Rightarrow F_n=0.0239364-0.175\times 2\times 10^{-3}\times 9.81-1.5\times 10^{-3}\times 9.81\\\Rightarrow F_n=0.0057879\ N$

The net vertical force is 0.0057879 N

6 0

3 years ago

The mechanical advantage of a wheel and axle is the radius of the wheel divided by the radius of the axle. TRUE or FALSE.

viva [34]

True! The mechanical advantage of the wheel and axle is equal to the ratio of the radius of the wheel over the radius of the axle.

4 0

3 years ago

What is the frequency of a pendulum of length 1.50 m at a location where 1 point the acceleration due to gravity is 9.79 m/s^2?

ivann1987 [24]

Answer:

Explanation:The simple pendulum calculator finds the period and frequency of a ... Acceleration of gravity (g) ... Pendulum length (L) ... First of all, a simple pendulum is defined to be a point mass or bob (taking ... For example, it can be equal to 2 m. ... Find the frequency as the reciprocal of the period: f = 1/T = 0.352 Hz

4 0

3 years ago

A person drives a car around a circular cloverleaf with a radius of 77 m at a uniform speed of 10 m/s. What is the acceleration

umka2103 [35]

Answer:

770m/s

Explanation:

caculation using one of the newton law of motion

6 0

3 years ago

Problem 1 Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed

ANTONII [103]

Answer:

markers are 29.76 m far apart in the laboratory

Explanation:

Given the data in the question;

speed of particle = 0.624c

lifetime = 159 ns = 1.59 × 10⁻⁷ s

we know that; c is speed of light which is equal to 3 × 10⁸ m/s

we know that

distance = vt

or s = ut

so we substitute

distance = 0.624c × 1.59 × 10⁻⁷ s

distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s

distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s

distance = 29.76 m

Therefore, markers are 29.76 m far apart in the laboratory

3 0

3 years ago