What are physical forms in which a substance can exist?

An automobile approaches a barrier at a speed of 20 m/s along a level road. The driver locks the brakes at a distance of 50 m fr

AleksAgata [21]

Answer:

μ = 0.408

Explanation:

given,

speed of the automobile (u)= 20 m/s

distance = 50 m

final velocity (v) = 0 m/s

kinetic friction = ?

we know that,

v² = u² + 2 a s

0 = 20² + 2 × a × 50

$a = \dfrac{400}{2\times 50}$

a = 4 m/s²

We know

F = ma = μN

ma = μ mg

a = μ g

$\mu = \dfrac{a}{g}$

$\mu = \dfrac{4}{9.81}$

μ = 0.408

hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408

5 0

2 years ago

A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$

$V(t) = 38 \frac{m}{s} - g * t$

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

$0 m = 38 \frac{m}{s} - g * t$

and obtain:

$t = 38 \frac{m}{s} / g$

$t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}$

$t = 3.9 s$

Plugin this time on our first equation we find:

$y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2$

$y=73.67 m$

6 0

2 years ago

A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleratio

dem82 [27]

Answer:

The torque on the child is now the same, τ.

Explanation:

It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:

$I_{to} = I_{d} + m*r^{2} = m*\frac{r^{2}}{2} + m*r^{2} = \frac{3}{2}* m*r^{2} (1)$

So, τ = 3/2*m*r²*α (2)

When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:

$I_{t} = I_{d} + m*\frac{r^{2}}{4} = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4} = \frac{3}{4}* m*r^{2} (3)$

Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:

τ = 3/4*m*r² * (2α) = 3/2*m*r²

same result than in (2), so the torque remains the same.

7 0

2 years ago

At a local swimming pool, the diving board is elevated h = 5.5 m above the pool's surface and overhangs the pool edge by L = 2 m

Margaret [11]

Answer:

Part a)

$t = \sqrt{\frac{2h}{g}}$

Part b)

$t = 1.06 s$

Part c)

$L = 4.86 m$

Explanation:

Part a)

The height of the diving board is given as

$h = 5.5 m$

now the speed of the diver is given as

$v_0 = 2.7 m/s$

when the diver will jump into the water then his displacement in vertical direction is same as that of height of diving board

So we will have

$y = v_y t + \frac{1}{2}at^2$

$h = 0 + \frac{1}{2}gt^2$

$t = \sqrt{\frac{2h}{g}}$

Part b)

$t = \sqrt{\frac{2h}{g}}$

plug in the values in the above equation

$t = \sqrt{\frac{2(5.5 m)}{9.81}$

$t = 1.06 s$

Part c)

Horizontal distance moved by the diver is given as

$d = v_0 t$

$d = 2.7 \times 1.06$

$d = 2.86 m$

so the distance from the edge of the pool is given as

$L = 2.86 + 2$

$L = 4.86 m$

4 0

2 years ago

Identical guns fire identical bullets horizontally at the same speed from the same height above level planes, one on the Earth a

Natasha_Volkova [10]

Answer:

I. The horizontal distance traveled by the bullet is greater for the Moon.

II. The flight time is less for the bullet on the Earth.

Explanation:

Horizontal distance depends on the initial speed, height and gravity. Bullets have the same initial speed and are shot from the same height. In these conditions horizontal distance only depends on gravity, which is inversely proportional. Therefore, the less gravity the greater the horizontal distance. Gravity slows bullet and causes its impact on the ground. Since gravity is greater in Earth, the bullet hits faster on the earth.

3 0

1 year ago

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