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3 years ago

Marcus drove his Honda Prelude for 4 hours at a rate of 55 miles per hour. How far did he travel?

2 answers:

7 0

To solve this, you’d multiply 55 by 4, because he is travelling 55 miles every hour, for four hours, which means 55 miles every hour. The answer would be 220.

lilavasa [31]3 years ago

5 0

Answer : Distance covered is 220 miles.

Explanation :

It is given that,

Speed of Marcus's Honda Prelude, $v=55\ miles/hour$

Time duration, $t=4\ hours$

We know that the speed of an object is defined as the distance travelled per unit time.

$v=\dfrac{d}{t}$

$d=v\times t$

$d=55\ miles/hour\times 4\ hours$

$d=220\ miles$

Hence, this is the required solution.

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What constant acceleration, in SI units, must a car have to go from zero to 60 mph in 10 s? How far has the car traveled when it

nalin [4]

Answer:

Explanation:

initial velocity, u = 0

final velocity, v = 60 mph = 26.8 m/s

time t = 10 s

Let a be the acceleration and s be he distance traveled.

Use first equation of motion

v = u + a t

26.8 = 0 + a x 10

a = 2.68 m/s

Use second equation of motion

s = ut + 1/2 at²

s = 0 + 0.5 x 2.68 x 10 x 10

s = 134 m

As, 1 m = 3.28 ft

So, s = 134 x 3.28 ft

s = 439.6 ft

7 0

3 years ago

A stubborn, 100 kgkg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around

jolli1 [7]

Answer:

No, the farmer is not able to move the mule.

Explanation:

Mass =100 kg

Force=F=800 N

The coefficient between the mule and the ground= $\mu_s=0.8$

$\mu_k=0.5$

Static friction force,f= $\mu N$

Normal force=N=mg

Static friction force,f= $\mu_s mg=0.8\times 100\times 9.8=784 N$

Using $g=9.8m/s^2$

F<f

Static friction force is greater than applied force.

Therefore , the farmer is not able to move the mule.

4 0

3 years ago

A 1.55-kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a 13.3-n hori

yarga [219]

<span>To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.

First, the vertical component of tension (Tsin theta) is equal to the weight of the object.
T * sin θ = mg =</span> 1.55 * 9.81 <span>
T * sin θ = 15.2055

Second, the horizontal component of tension (t cos theta) is equal to the force of the wind.
T * cos θ = 13.3

Tan θ = sin </span>θ / cos θ = 15.2055/13.3 = 1.143
we can find θ that is equal to 48.82.

T then is equal to 20.20 N

4 0

3 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$