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aniked [119]
3 years ago
15

A car went 60 km in 5/6 of an hour while a second car went 54 km in 2/3h. Which car was faster? How many times faster?

Physics
1 answer:
Nastasia [14]3 years ago
4 0

Answer:

The car that went 54 km in 2/3h was faster, 5/4 times faster than the other car

Explanation:

Average speed of a car is the ratio between the displacement \Delta x and the time (t) it takes to do that displacement:

V_{avg}=\frac{\Delta x}{t}

So, for the first car:

V_{avg1}=\frac{60km}{\frac{5h}{6}} =72\frac{km}{h} (1)

for the second car we have:

V_{avg2}=\frac{60km}{\frac{2h}{3}} =90\frac{km}{h} (2)

So, the second cart is faster than the first one. To find how many times divide speed 2 by speed 1:

\frac{90}{72}=\frac{5}{4}

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A soccer ball is released from rest at the top of a grassy incline. After 6.2 seconds, the ball travels 47 meters. One second la
alexandr1967 [171]

Answer:

(a) a = 2.44 m/s²

(b) s = 63.24 m

Explanation:

(a)

We will use the second equation of motion here:

s = v_it+\frac{1}{2}at^2

where,

s = distance covered = 47 m

vi = initial speed = 0 m/s

t = time taken = 6.2 s

a = acceleration = ?

Therefore,

47\ m = (0\ m/s)(6.2\ s)+\frac{1}{2}a(6.2\ s)^2\\\\a = \frac{2(47\ m)}{(6.2\ s)^2}

<u>a = 2.44 m/s²</u>

<u></u>

(b)

Now, we will again use the second equation of motion for the complete length of the inclined plane:

s = v_it+\frac{1}{2}at^2

where,

s = distance covered = ?

vi = initial speed = 0 m/s

t = time taken = 7.2 s

a = acceleration = 2.44 m/s²

Therefore,

s = (0\ m/s)(6.2\ s)+\frac{1}{2}(2.44\ m/s^2)(7.2\ s)^2\\\\

<u>s = 63.24 m</u>

6 0
3 years ago
A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to acc
Jlenok [28]

Answer:

a is the answer

Explanation:

3 0
3 years ago
Read 2 more answers
The distance between planets varies as they move in their orbits. The minimum distance from the Earth to Mars is about 0.35 AU.
Alina [70]

Answer:

52,360,000km

Explanation:

To solve this problem you use a conversion factor.

By taking into account that 1UA = 1.496*10^{8}km you obtain:

0.35UA*\frac{1.496*10^{8}km}{1UA}=52,360,000km

hence, 0.35UA is about 52,360,000km. This is the least distance between Mars and Earth

5 0
3 years ago
Read 2 more answers
A person walks 2 miles every day to work, leaving her front porch at 7:00 A.M. and arriving at work at 7:30 A.M. ON the way home
Nataly [62]

Answer:

The displacement is zero miles

Explanation:

The displacement of an object that moves from point A to point B is defined as

d =B-A

Where d is the displacement of the object. The displacement does not depend on the trajectory of the object. It only depends on the linear distance between the end point and the starting point.

In this case we know that the person walks from home to work and then walks from work to home. Therefore, the total displacement is the linear distance between the point where its journey begins and the point where the route ends.

The tour begins on the front porch of your house and ends on the front porch of your house (when you return from work). If we call A to the front porch of the house then the displacement is:

d = A-A = 0

The displacement is zero miles, since the person finishes the journey just where it started (front porch)

7 0
3 years ago
g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll
irga5000 [103]

Answer:

The minimum coefficient of static friction required, µ = 0.10

<em>Note. The question is incomplete. The complete question is given below:</em>

<em>While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.</em>

<em>The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.</em>

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10

8 0
2 years ago
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