Label the samples based on their pH levels

Suppose you toss a tennis ball upward.a) Does the kinetic energy of the ball increase or decrease as it moves higher?b) What hap

Lilit [14]

Answer:

a) No. The kinetic energy of the ball decreases.

b) The potential energy of the ball increases.

c) The ball would go half of the original distance.

Explanation:

a) The kinetic energy would be converted to potential energy as the ball goes higher. Since the total mechanical energy is conserved, the kinetic energy would decrease.

b) The potential energy of the ball would increase. Since the total mechanical energy of the ball is conserved, the ball would lose speed, and therefore kinetic energy. In order to compensate the loss of kinetic energy, the ball would gain potential energy as it goes higher.

c) The relation of the energy and mass is as follows:

$K = \frac{1}{2}mv^2\\U = mgh$

According to the energy conservation

$K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv^2 + 0 = 0 + mgh\\\frac{1}{2}v^2 = gh$

The maximum height that the ball reaches is proportional to the initial velocity. If the ball would be imparted with the same amount of energy, its final potential energy would be the same. However, in order to have the same potential energy (mgh), its height would be half of the original case.

$mgh = (2m)g(\frac{h}{2})$

7 0

3 years ago

A car has a mass of 1520 kg. While traveling at 20 m⁄s, the driver applies the brakes to stop the car on a wet surface with a 0.

docker41 [41]

Answer:

(a)d₁ = 51.02 m

(b)d₂ =51.02m

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

Known data

m=1520 kg : mass of the car

μk= 0.4 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the car

We define the x-axis in the direction parallel to the movement of the car and the y-axis in the direction perpendicular to it.

W: Weight of the block : In vertical direction downward

FN : Normal force : In vertical direction upward

f : Friction force: In horizontal direction

Calculated of the W

W= m*g

W= 1520 kg* 9.8 m/s² = 14896 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay ay = 0

FN - Wy = 0

FN = Wy

FN = 14896 N

Calculated of the f

f = μk* N= (0.4)* (14896 N )

f = 5958.4 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

- f = m*a

-5958.4 = (1520)*a

a = (-5958.4) / ((1520)

a = -3.92 m/s²

(a) displacement of the car (d₁)

Because the car moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d₁ Formula (2)

Where:

d:displacement (m)

v₀: initial speed (m/s)

vf: final speed (m/s)

Data:

v₀ = 20 m⁄s

vf = 0

a = --3.92 m/s²

We replace data in the formula (2) to calculate the distance along the ramp the block reaches before stopping (d₁)

vf²=v₀²+2*a*d ₁

0 = (20)²+2*(-3.92)*d ₁

2*(3.92)*d₁ = (20)²

d₁ = (20)² / (7.84)

d₁ = 51.02 m

(b) Different car

m₂ = 1.5 *1520 kg

μk₂= 0.4

W₂= m*g

W₂= (1.5) *1520 kg* 9.8 m/s² = (1.5)*14896 N

FN₂= (1.5)*14896 N

f= 0.4* (1.5)*14896 N

a = - f/m₂ = - 0.4* (1.5)*14896 N /(1.5) *1520

a = -3.92 m/s²

vf²=v₀²+2*a*d₂

vf=0 , v₀=20 m⁄s , a = -3.92 m/s²

d₂ = d₁ = 51.02m

6 0

4 years ago

A child on a swing falls a height of 1m. What speed will the child reach?

kirill [66]

Answer: $v=4.42\ m/s$

Explanation:

Given

Child falls a height of $h=1\ m$

Applying conservation of energy we can write

Loss in Potential energy=Gain in kinetic energy

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

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 1}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

Thus child reaches a velocity of 4.42 m/s

7 0

4 years ago

In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres o

SVETLANKA909090 [29]

Answer:

$F = 7.7*10^{-10}N$

Explanation:

You need to be careful with units for this problem. The force will be:

$F =\frac{K*m1*m2}{d^2}$

$F=\frac{6.67259 * 10^{-11}*1.56*21.1*10^{-3}}{(5.34*10^{-2})^2}$

$F=7.7*10^{-10}N$

8 0

3 years ago

The sound intensity of a certain type of food processor in normally distributed with standard deviation of 2.9 decibels. If the

Maru [420]

Answer: $(48.41,\ 52.19)$

Explanation:

The confidence interval for population mean is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : $n=9$

Sample mean : $\ovreline{x}=50.3\text{ decibels}$

Standard deviation : $\sigma=2.9\text{ decibels }$

Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=z_{0.025}=1.96$

Now, the 95% confidence interval estimate of the (true, unknown) mean sound intensity of all food processors of this type :-

$50.3\pm (1.96)\dfrac{2.9}{\sqrt{9}}\\\\\approx50.3\pm1.89\\\\=(50.3-1.89,\ 50.3+1.89)=(48.41,\ 52.19)$

6 0

3 years ago

Label the samples based on their pH levels​

Label the samples based on their pH levels