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Nataly_w [17]
4 years ago
14

The buoyant force on an object is least when the object is

Physics
1 answer:
Arte-miy333 [17]4 years ago
6 0
When the object is partially submerged
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Could anyone help me on the questions of the Series Circuits please!!!! (The picture is attached)
strojnjashka [21]
Series circuits - cpt connected so that the same current passes through each of them. 
Disconnection/break in series circuit and current doesn't flow.
Current same 
The total potential difference across all of the components. (the sum of the pds across the individual series elements).

The diagram in the enclosure seems to show resistors in parallel. Which is a bit confusing ...

6 0
3 years ago
A boy rows a rowboat across a river. The boat moves at 4.3 m/s at a direction of 25°
NikAS [45]

East component: 3.9 m/s

South component: 1.8 m/s

Explanation:

We have to resolve the velocity vector along the east and south axis.

Taking east as positive x-direction and south as positive y-direction, the components of the velocity are given by:

v_x = v cos \theta\\v_y = v sin \theta

where

v = 4.3 m/s is the magnitude of the velocity

\theta=25^{\circ} is the angle between the direction of the velocity and of the x-axis

Substituting into the equations, we find:

East component:

v_x = (4.3)(cos 25^{\circ})=3.9 m/s

South component:

v_y = (4.3)(sin 25^{\circ})=1.8 m/s

Learn more about vector components:

brainly.com/question/2678571

#LearnwithBrainly

5 0
4 years ago
Which two conditions explain why special technology is required to study the
enot [183]

Answer:

A. Extremely high pressure

5 0
2 years ago
Read 2 more answers
A car stops in 120 m. If it has an acceleration of –5m/s 2 , how long did it take to stop
vova2212 [387]

Answer:

t=240s

Explanation:

Distance=120m

Acceleration=-5m/s^2

v=0

Let u=x m/s

Using equation v^2-u^2=2as:-

0-x=2(-5)(120)

-x=-1200

x=1200m/s

Using now equation v=u+at:-

0=1200+-5t

5t=1200

t=240s

8 0
4 years ago
A truck is moving at 25.0 m/s and sees a barrier in the road 100 m ahead. The truck can decelerate at 4 m/s2. Will the truck hit
Nimfa-mama [501]

Answer:

No, the truck will not cross the barrier.

The closeness of the truck to the barrier is of 21.875 m

Solution:

As per the question:

Velocity of the truck, v = 25.0 m/s

Acceleration of the truck, a = - 4 m/s^{2}

Now,

Since, the barrier at a distance of 100 m. Thus in order to check whether the truck hit the barrier or not, we will see the distance, d it covers by using the kinematic eqn:

v'^{2} = v^{2} + 2ad

Final velocity, v' = 0 m/s

Initial velocity = v

Now,

0^{2} = 25^{2} + 2\times -4d

- 8d = - 625

d = 78.125 m

Thus the truck will not cross the barrier.

Distance between the barrier and the truck:

100 - 78.125 = 21.875 m

6 0
4 years ago
Read 2 more answers
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