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AURORKA [14]
2 years ago
11

Winner of five swimming gold medals at the 1988 olympics

Physics
1 answer:
Fed [463]2 years ago
6 0
Matt Biondi..?
(I don’t know if it’s right, sorry if it is wrong)
:)
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Two equal resistors are connected in series with a 1.50V battery. In order to keep the current at 0.030 A, the resistors much ea
AlexFokin [52]

Answer: 25 Ohms

Explanation:

From this question, the following parameters are given:

Voltage V = 1.5 v

Current I = 0.03A

From Ohm's law;

V = IR

Where R = resultant resistance of the two resistors.

Substitute V and I into the formula and make resultant R the subject of formula.

1.5 = 0.03 × R

R = 1.5/0.03

R = 50 Ohms

From the question, it is given that Thr two equal resistors are connected in series.

R = R1 + R2

But R1 = R2

50 = 2R1

R1 = 50/2

R1 = 25

R1 = R2 = 25 Ohms

Therefore, the resistors must each have a value of 25 Ohms

4 0
3 years ago
10. How much total work do you do when you lift a 50 kg microwave 1.0 m off the ground and then push it 1.0 m
Mariulka [41]

Work formula:

W = Fd\cos(\theta)

F = 50N, d = 1.0 m

When you lift something straight up, the angle of the force is 90º

cos(90º) is 0, so there's no work done when you lift the microwave off the ground

W = (50N)(1.0)(0) = 0

F = 50N, d = 1.0 m

When you push the microwave, the angle is 0º and cos(0º) is 1. So there is work done here:

W = (50 N)(1.0m)(1)

W = 50

total work = 50 joules

6 0
2 years ago
How does the law of conservation of mass apply to this reaction: C2H4 + O2 → H2O + CO2?
kicyunya [14]

The answer is a

the equation needs to be balanced. There are fewer oxygen atoms in the equation than hydrogen or a carbon

4 0
3 years ago
Read 2 more answers
In the figure below the pulley is a solid disk of mass M and radius R with rotational inertia MR 2/2. Two blocks one of mass m a
matrenka [14]
Assuming you are looking for the acceleration a:

1.m_1a = T_1 -m_1g
2.m_2a = m_2g - T_2
where T is the tension and a is the acceleration of the blocks. The acceleration of the two blocks and the acceleration of the pulley must be equal.

The torque on the pulley is given by:
3.\tau = \overrightarrow r \times \overrightarrow F = (T_2 - T_1)R = I\alpha = \frac{1}{2} MR^2 \frac{a}{R}
where I = \frac{1}{2} mR^2 and a = \alpha R.

Combining the three equations:
T_2 - T_1 = \frac{1}{2} Ma \\ m_2g - m_2a -m_1g - m_1a = (m_2-m_1)g - (m_1 + m_2)a = \frac{1}2}Ma \\ \\ a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2}M }
6 0
3 years ago
An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the
OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2

Through the aforementioned formula we will have to

v_f^2-v_i^2 = 2ax

The particulate part of the rest, so the final speed would be

v_f^2 = 2gx

v_f=\sqrt{2(9.8)(3.1)}

v_f = 7.79m/s

Now from Newton's second law we know that

F = ma

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

F = m\frac{dv}{dt}

Replacing we have that,

F = (89)\frac{7.79}{0.5}

F = 1386.62N

Therefore the force that the water exert on the man is 1386.62

3 0
3 years ago
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