A bucket of water with weight of 50N is pulled up using a rope. The bucket is moving at

Câu 1. Con lắc lò xo treo thẳng đứng, dao động điều hòa với biên độ 2cm và tần số góc 20 rad/s. Chiều dài tự nhiên của lò xo là

natka813 [3]

Answer:

28 cm and 32 cm

Explanation:

1. The spring pendulum hangs vertically, oscillates harmonic with amplitude 2cm and angular frequency 20 rad/s. The natural length of

a spring is 30cm. What is the minimum and maximum length of the spring during the oscillation? Take g = 10m/s2.

As the amplitude is 2 cm and the natural length is 30 cm. So, it oscillates between 30 -2 = 28 cm to 30 + 2 = 32 cm.

So, the minimum length is 28 cm and the maximum length is 32 cm.

8 0

2 years ago

A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N. The magnitude of the electric f

zalisa [80]

Explanation:

Charge=5.00 E-5

Force=0.751N

F=qE

0.751=5.00 E-5*E

E=1.502*10⁴

option a

8 0

2 years ago

Read 2 more answers

Can you help me??????????????????????????)?

garik1379 [7]

Answer:

Velocity is a change in displacement over change in time and uses the units m/s.

Both are rates of change and can be positive or negative.

Acceleration is a change in velocity over change in time and uses the units m/s².

Explanation:

Velocity is the change in displacement over change in time, this makes it a rate of change. It can be positive or negative because it is a vector quantity. It uses the units m/s because that is a displacement unit over a time unit.

Acceleration is the change in velocity over change in time, this makes it a rate of change. It can be positive or negative because it is also a vector quantity. It uses the units m/s² (m/s/s) because that is a velocity unit over a time unit.

5 0

2 years ago

I was driving along at 20 m/s, trying to change a CD and not watching where I was going. When I looked up, I found myself 45 m f

cricket20 [7]

Answer:

$a=4.44\frac{m}{s^2}$

Explanation:

First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

$t=\frac{d}{v}\\\\t=\frac{60m}{30\frac{m}{s}}=2s$

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.

Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

$d=vt\\d=20\frac{m}{s}(0.5s)=10m$

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:

$d=v_0t+\frac{1}{2}at^2\\a=\frac{2(d-v_0t)}{t^2}\\a=\frac{2(35m-20\frac{m}{s}*1.5s}{(1.5s)^2}\\a=4.44\frac{m}{s^2}$

7 0

2 years ago

What are the genotypes of gametes of a aabb self-pollination??

Rus_ich [418]

All of the possibilities would be aabb. There are no dominant traits to come through from the self pollinating plant so therefore, the offspring would would have the same genotypes.

I apologize for the messy graph in advance. If you have any further questions please let me know

5 0

2 years ago