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3 years ago

A car with a velocity of 15 m/s is accelerated uniformly at the rate of 1.9 m/s2 for 6.8 s. What is its final velocity?

Physics

1 answer:

creativ13 [48]3 years ago

3 0

Answer:

33 m/s

Explanation:

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A simple machine that is a straight slanted surface is

rjkz [21]

Inclined plane is a simple machine that is a straight slanted surface.

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3 years ago

The average velocity of an object over 6.0 seconds interval is 2 m/s what is the total distance traveled and M by the object doi

lina2011 [118]

Answer:

<h2>The answer is 12 m</h2>

Explanation:

The distance covered by an object given it's velocity and time taken can be found by using the formula

distance = velocity × time

From the question we have

distance = 2 × 6

We have the final answer as

Hope this helps you

8 0

2 years ago

A Shaolin monk of mass 60 kg is able to do a ‘finger stand’: he supports his whole weight on his two index fingers, giving him a

LekaFEV [45]

Answer:

1500000 Pa

Explanation:

The formula for pressure is force per unit area.

P=F/A where F is force and A is area

Given that ;

F= mass * acceleration due to gravity

F= 60 * 9.81 = 588.6 = 589 N

A= area = 4cm² = 0.0004 m²

P= F/A = 589 / 0.0004

P= 1471500

P=1500000 Pa

4 0

3 years ago

Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed

olasank [31]

Answer:

Average acceleration is $(11.05)g\ m/s^2$

Explanation:

It is given that,

Initial velocity, u = 0

Final velocity, v = 6.5 km/s = 6500 m/s

Time taken, t = 60 s

Acceleration, $a=\dfrac{v-u}{t}$

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

$a=\dfrac{6500\ m/s}{60}$

$a=108.33\ m/s^2$

Since, $g=9.8\ m/s^2$

So, $a=(11.05)g\ m/s^2$

So, the angular acceleration of the missile is $(11.05)g\ m/s^2$ . Hence, this is the required solution.

4 0

3 years ago

Qual a capacidade térmica de um objeto que, ao receber 10000 cal de energia, tem sua temperatura elevada de 25°C para 75°C?

Stolb23 [73]

Answer:

$200 cal/^{\circ}C$

Explanation:

When heat energy is supplied to an object, the temperature of the object increases according to the equation:

$Q=C\Delta T$

where

Q is the heat supplied

C is the heat capacity of the object

$\Delta T$ is the change in temperature

In this problem we have:

$Q=10,000 cal$ is the energy supplied

$\Delta T=75C-25C=50C$ is the change in temperature of the object

Therefore, the heat capacity of the object is:

$C=\frac{Q}{\Delta T}=\frac{10,000}{50}=200 cal/^{\circ}C$

6 0

3 years ago