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UNO [17]
2 years ago
11

A car is traveling north at 17.7m/s After 12 s its velocity is 14.1m/s in the same direction. Find the magnitude and direction o

f the car's average acceleration.
a2.7 m/s2, north b0.30 m/s2, south c2.7 m/s2, south d0.30 m/s2, north
Physics
1 answer:
Nat2105 [25]2 years ago
7 0

The car is initially traveling north at 17.7 m/s, and after 12 s, its velocity is 14.1 m/s, still due north. This means that the direction of the car has not changed, so we can already say that the direction of the acceleration is north (if the magnitude of the acceleration is positive) or south (if the magnitude of the acceleration is negative).

To find the magnitude of the average acceleration, we must calculate the ratio between the change in velocity and the time taken:

a=\frac{v_f -v_i}{t}=\frac{14.1 m/s-17.7 m/s}{12 s}=-0.3 m/s^2

Since the acceleration is negative, it means it is in the opposite direction to the motion of the car, therefore south. Therefore, the correct answer is

b) 0.30 m/s2, south

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You use a slingshot to launch a potato horizontally from the edge of a cliff with speed v0. The acceleration due to gravity is g
Ray Of Light [21]

Answer:

\displaystyle t=\frac{2v_o}{g}

Explanation:

<u>Horizontal Launch</u>

When an object is launched horizontally at a speed vo, it describes a curved called parabola as the speed in the x-direction does not change and the speed in the y-direction increases with time because the gravity makes it return to the ground.

The vertical distance the object (potato) travels downwards is:

\displaystyle y=\frac{gt^2}{2}

The horizontal distance is

x=v_ot

We need to find the time when both distances are equal, thus

\displaystyle \frac{gt^2}{2}=v_ot

Simplifying by t

\displaystyle \frac{gt}{2}=v_o

Solving for t

\displaystyle \boxed{t=\frac{2v_o}{g}}

8 0
2 years ago
on the surface of planet x a body with a mass of 10 kilograms weighs 40 newtons. The magnitude of the acceleration due to gravit
tamaranim1 [39]
Based on the Newton's second law of motion, the value of the net force acting on the object is equal to the product of the mass and the acceleration due to gravity. If we let a be the acceleration due to gravity, the equation that would allow us to calculate it's value is,
      W = m x a
where W is weight, m is mass, and a is acceleration. Substituting the known values,
    40 kg m/s² = (10 kg) x a
Calculating for the value of a from the equation will give us an answer equal to 4. 
ANSWER: 4 m/s². 

6 0
2 years ago
Read 2 more answers
In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
SpyIntel [72]

Answer:

A) F_g = 26284.48 N

B) v = 7404.18 m/s

C) E = 19.19 × 10^(10) J

Explanation:

We are given;

Mass of satellite; m = 3500 kg

Mass of the earth; M = 6 x 10²⁴ Kg

Earth circular orbit radius; R = 7.3 x 10⁶ m

A) Formula for the gravitational force is;

F_g = GmM/r²

Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Plugging in the relevant values, we have;

F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²

F_g = 26284.48 N

B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.

Thus;

GmM/r² = mv²/r

Making v th subject, we have;

v = √(GM/r)

Plugging in the relevant values;

v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))

v = 7404.18 m/s

C) From the energy principle, the minimum amount of work is given by;

E = (GmM/r) - ½mv²

Plugging in the relevant values;

E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)

E = 19.19 × 10^(10) J

5 0
2 years ago
A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f
sasho [114]

We use the kinematic equations,

v=u+at                                          (A)

S= ut + \frac{1}{2} at^2                  (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given,  u=0, a=0.21 \ m/s^2 and s= 280 m.

Substituting these values in equation (B), we get

280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s.

Therefore from equation (A),

v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s



8 0
2 years ago
Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and
Anna11 [10]

Kepler’s third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496 × 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/( 86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The closest answer is 1.99 × 10^30

(it may vary a little with rounding – the difference is less than 1%)

5 0
2 years ago
Read 2 more answers
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