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Anit [1.1K]
3 years ago
13

The current, 2019, women's world record for the

Physics
1 answer:
sesenic [268]3 years ago
8 0

Answer: D

8.403 m/s

Explanation:

Given that the women's world record for the 400-meter run around an oval track is held by Marita Koch, with a time of 47.60

seconds.

Her average velocity during this race will be

Velocity = distance/ time

Velocity = 400 / 47.6

Velocity = 8.403 m/s

Therefore, her average velocity during this race will be 8.403 m/s

D is therefore the correct answer.

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Why sheets of newspaper can be used to keep ice cream cold and keep fish and chips hot?
stich3 [128]

Answer: Paper

Explanation: It is a insulator

7 0
4 years ago
Which statement is true? CTP hydrolysis can be a source of chemical energy. dATP is the most common source of chemical energy in
Gemiola [76]

1. Statement A is true.

When GTP is hydrolysed, the free energy of hydrolyses is used to power or drive reactions that are favourable energetically.

2. Statement B it true.

ATP is a complex chemical that gives energy for the activities in many living cells. During hydrolyses, chemical energy stored in the energy-rich phosphoanhydride is released. Hence its a common source of chemical energy in cells.

3. Statement C is false.

The hydrolyses of ATP to ADP in the presence of phosphate, releases one mole of ATP which is estimated to be -57Kj/mol not 14Kj/mol. Below is the equation;

ATP + H20 -----> ADP + Pi + Free energy.

4. Statement D is false.

GTP stands for guanosine triphosphate.

ATP stands for adenosine triphosphate

ADP stands for adenosine diphosphate.

3 0
4 years ago
A mass m = 4.6 kg hangs on the end of a massless rope L = 2.16 m long. The pendulum is held horizontal and released from rest.
serious [3.7K]

ind the velocity my using the conservation of energy mgh=1/2mv^2. At the bottom, the mass is turning in a circle with tension pointing up towards the center of the circle and weight pointing down. Use the centripetal force requirement mv^2/r=T-W and solve for T. Hope this helps.

4 0
3 years ago
Find the weight required to stretch a steel road by 2 mm, if the length of the road is 2 m and diameter 4 mm. Y= 200 GPa and g -
svetlana [45]

Answer:

Option D is the correct answer.

Explanation:

We equation for elongation

   \Delta L=\frac{PL}{AE}      

Here we need to find weight required,

We need to stretch a steel road by 2 mm, that is ΔL = 2mm = 0.002 m

A=\frac{\pi d^2}{4}=\frac{\pi ({4\times 10^{-3}})^2}{4}=1.26\times 10^{-5}m^2

E = 200GPa = 2 x 10¹¹ N/m²

L=2m

Substituting

0.002=\frac{P\times 2}{1.26\times 10^{-5}\times 2\times 10^{11}}\\\\P=2520N=256.4kg

Option D is the correct answer.

6 0
3 years ago
You are working in cooperation with the Public Health department to design an electrostatic trap for particles from auto emissio
Mademuasel [1]

Answer:

Explanation:

Given that:

Charge (q) on the particle = 3 × 10⁻⁸ C

mass (m) of the particle = 6 × 10⁻⁹ kg

at a distance x = 15 cm , the velocity in the plate = 900 m/s²

For the square plate, the surface charged density σ = -8 × 10⁻⁶ C/m²

To start with calculating the electric field as a result of the square plate; we use the formula;

E = \dfrac{\sigma }{2 \varepsilon_o}

E = \dfrac{8 \times 10^{-6} }{2 \times  8.85 \times 10^{-12}}

E = 4.51977 \times 10^5 \ V/m

On the square plate; The electric force F = Eq

F = (4.51977 \times 10^5 \ V/m )(3\times 10^{-8} \ C)

F = 1.3559 \times 10^{-2} \ N

The acceleration a =\dfrac{ F}{m}{

a = \dfrac{1.3559\times 10^{-2} \ N}{6 \times 10^{-9} \ Kg}

a = 2.25988 \times 10^6 \ m/s^2

For the particle, the velocity at distance x = 7 m can be calculated by using the formula:

(\dfrac{1}{2}) mv^2 = \Delta Vq

v^2 = \dfrac{2 Eq}{dm}

v^2 = \dfrac{2 * 4.51977 \times 10^5 \times 3 \times 10^{-8} }{0.07 \times 6\times 10^{-9} }

v^2 = 64568142.86  \ m/s

v =\sqrt{ 64568142.86  \ m/s}

\mathbf{v = 8.035 \times 10^3 \ m/s}

From the calculation, we realize that the charge acting between the particle and the plate is said to be "opposite".

Hence, the force is an attractive force.

Similarly, there is a gradual increase exhibited by the velocity of the particle.

Therefore, the particles get to the detector, but the detector failed to get detect due to the velocity which is greater than 1000 m/s.

6 0
3 years ago
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