A model train moves 18.3m in 122s. What is the train's average speed?

A cart is set up as shown below, with three fans directed to the left and two of the fans running. The motion of the cart is rep

sasho [114]

The force of 3 fans would be grater than the force of 2 fans.
So the cart would have greater acceleration to the right, and
it would reach a greater speed to the right. The resulting
'v' vs 't' graph would have a greater positive slope, and would
be higher at any given time.

You might get a better answer if you let us actually SEE the
picture and the graph.

5 0

4 years ago

Read 2 more answers

If a ball is thrown upward with an initial velocity of 30.0 m/s how much time does it take to reach its maximum height

Masteriza [31]

Gravity makes an upward-moving object move 9.8 m/s slower every second.

(30 m/s) / (9.8 m/s²) = 3.06 seconds (rounded)

3 0

3 years ago

What is the mass of a baseball that has a kinetic energy of 105 J and is traveling at 10 m/s?

kkurt [141]

Answer:

$\boxed {\boxed {\sf 2.1 \ kilograms}}$

Explanation:

Kinetic energy is the energy an object possesses due to motion. The formula 1/2 the product of mass and the squared velocity.

$E_k=\frac{1}{2} mv^2$

We know the baseball's kinetic energy is 105 Joules. It is also traveling at a velocity of 10 meters per second. `

First, convert the units of Joules to make unit cancellation easier later in the problem. 1 Joule (J) is equal to 1 kilogram square meter per square second (kg*m²/s²). The baseball's kinetic energy of 105 J is equal to 105 kg*m²/s².

Now we know 2 values:

$E_k= 105 \ kg*m^2/s^2$

$v= 10 \ m/s$

Substitute these values into the formula.

$105 \ kg*m^2/s^2= \frac{1}{2} m (10 \ m/s)^2$

Now we need to solve for m, the mass. Solve the exponent.

(10 m/s)²= 10 m/s * 10 m/s = 100 m²/s²

$105 \ kg *m^2/s^2 = \frac{1}{2} m (100 \ m^2/s^2)$

Multiply on the right side.

$105 \ kg *m^2/s^2 = m (\frac{1}{2} * 100 \ m^2/s^2)$

$105 \ kg *m^2/s^2 = m (50 \ m^2/s^2)$

The variable, m, is being multiplied by 50 square meters per square second. The opposite of multiplication is division, so we divide both sides by that value.

$\frac {105 \ kg *m^2/s^2 }{50 \ m^2/s^2}= \frac{ m (50 \ m^2/s^2)}{50 \ m^2/s^2}$

$\frac {105 \ kg *m^2/s^2 }{50 \ m^2/s^2}= m$

The units of square meter per square second will cancel out.

$\frac {105 }{50} \ kg= m$

$2.1 \ kg=m$

The mass of the baseball is <u>2.1 kilograms. </u>

5 0

3 years ago

Two uniform bars of the same dimensions are constructed from the same material. One bar has five evenly spaced holes through it

Elan Coil [88]

Solution :

The angular acceleration, $\alpha$ is obtained from the equation of the $\text{Newton's second law}$ of rotational motion,

Thus,

$\tau = F \times d$

or $\tau = I \times \alpha$

where $\tau$ is torque, F is force, d is moment arm distance, I is the moment of inertia

Thus, $\alpha=\frac{(F\times d)}{I}$

Now if the force and the moment arm distance are constant, then the $\text{angular acceleration is inversely proportional to the moment of inertia.}$

That is when, F = d = constant, then $\alpha \propto \frac{1}{I}$ .

Thus, moment of inertia, I is proportional to mass of the bar.

The mass is less for the bar in case (1) in comparison with that with the bar in case (2) due to the holes that is made in the bar.

Therefore, the bar in case (1), has less moment of inertia and a greater angular acceleration.

6 0

3 years ago

ASAP please help me

just olya [345]

Answer:

1.2 s

Explanation:

We'll begin by calculating the length (i.e distance) of the ramp. This can be obtained by using pythagoras theory as illustrated below:

NOTE: Length of the ramp is the Hypothenus i.e the longest side.

Let the Lenght of the ramp be 's'. The value of x can be obtained as follow:

s² = 4² + 3²

s² = 16 + 9

s² = 25

Take the square root of both side

s = √25

s = 5 m

Thus the length of the ramp is 5 m

Next, we shall determine the final velocity of the ball. This can be obtained as follow:

Initial velocity (u) = 3 m/s

Acceleration (a) = 2 m/s²

Distance (s) = 5 m

Final velocity (v) =?

v² = u² + 2as

v² = 3² + (2 × 2 × 5)

v² = 9 + 20

v² = 29

Take the square root of both side

v = √29

v = 5.39 m/s

Finally, we shall determine the time taken for the ball to reach the final position. This can be obtained as follow:

Initial velocity (u) = 3 m/s

Acceleration (a) = 2 m/s²

Final velocity (v) = 5.39 m/s

Time (t) =?

v = u + at

5.39 = 3 + 2t

Collect like terms

5.39 – 3 = 2t

2.39 = 2t

Divide both side by 2

t = 2.39 / 2

t = 1.2 s

Thus, it will take 1.2 s for the ball to get to the final position.

4 0

3 years ago