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adelina 88 [10]
4 years ago
8

If a truck is travelling east on a straight road and travels 100 meters in 25s what is the truck's velocity?

Physics
1 answer:
Slav-nsk [51]4 years ago
3 0

Answer:

Velocity of the truck, v = 4 m/s                                                            

Explanation:

It is given that,

Distance covered by the truck, d = 100 m

Time taken by the truck, t = 25 s

Let v is the velocity of the truck. We know that the velocity is a vector quantity. Mathematically, it is given by :

v=\dfrac{d}{t}

v=\dfrac{100\ m}{25\ s}

v = 4 m/s

So, the velocity of the truck is 4 m/s. Hence, this is the required solution.

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A car in an amusement park ride rolls without friction around a track (Fig. P7.42). The hB car starts from rest at point A at a
MrRissso [65]

Answer:

h>\dfrac{5}{2}R

Explanation:

Given that

Height = h

Radius = R

From energy conservation

KE_A+U_A=KE_B+U_B

At point B

The minimum speed to complete the   the circle

V_B=\sqrt{gR}\ m/s

So the kinetic energy at point B

KE_B=\dfrac{1}{2}mV^2

KE_B=\dfrac{1}{2}mgR

KE_A+U_A=KE_B+U_B

0+mgh=\dfrac{1}{2}mgR+2mgR

Without falling off at the top (point B)

0+mgh>\dfrac{1}{2}mgR+2mgR

mg(h-2R)>\dfrac{1}{2}mgR

g(h-2R)>\dfrac{1}{2}gR

h>\dfrac{5}{2}R

6 0
4 years ago
A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.
Airida [17]

Answer:

a) v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} ), b) \Delta K = 9.559\times 10^{-5}\,J

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v

The final velocity is:

v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )

b) The change in the kinetic energy of the 13.5 g coin is:

\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]

\Delta K = 9.559\times 10^{-5}\,J

4 0
3 years ago
Read 2 more answers
A 2.00-kg ball is moving at 2.20 m/s toward the right. It collides elastically with a 4.00-kg ball that is initially at rest. 1)
loris [4]
Elastic collision is when kinetic energy before = kinetic energy after

Ek= 1/2mv^2

total before
Ek=1/2(2)(2.2^2) = 4.84 J

total after
Ek= 1/2(2+4)(v^2) = 3v^2

Before = after
4.84=3v^2 | divide by 3
121/75 = v^2 | square root both sides
v=1.27 m/s
7 0
3 years ago
Before using a string in a comparison, you can use either the To Upper method or the To Lower method to convert the string to up
diamong [38]

Answer:

True, check attachment for code

Explanation:

To convert java strings of text to upper or lower case, we can use and inbuilt methods To Uppercase and To lower case.

The first two lines of code will set up a String variable to hold the text "text to change", and then we print it out.

The third line sets of a second String variable called result.

The fourth line is where the conversion is done.

We can compare the string

We can compare one string to another. (When comparing, Java will use the hexadecimal values rather than the letters themselves.) For example, if we wanted to compare the word "Fat" with the word "App" to see which should come first, you can use an inbuilt string method called compareTo.

Check attachment for the code

3 0
3 years ago
How does the transformation of energy takes place in crackers
iragen [17]
Your answer is Chemical to Thermal and Electromagnetic c:
4 0
3 years ago
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