How much current does a 10.0 Ω resistor draw from a 12 V battery?

A motorcycle is traveling east at a rate of 56mph.it take 6.7 s for it to decrease its speed to 35mph. How far does the motorcyc

Phantasy [73]

Initial speed = 56mph
Final speed = 35mph

Time taken = 6.7seconds...

Converting the time to hour.. Divide by 3600..
= 6.7/3600

=0.00186hour..

Acceleration = v-u/t

a = 35-56/0.00186
a = -11283.6mph²

The negative sign shows that it decelerated...

V² = u²+2as

(35)² = (56)² + 2×-11283.6×s
Where s is the distance covered within that time...

1225 = 3136 - 22567.2s

22567.2s = 3136-1225

22567.2s = 1911

S = 1911/22567.2

S = 0.08468miles...

But at the end of the question we were made to understand that 1miles = 5280ft

Therefore 0.08468miles = (0.08468×5280)ft

= 447. 11feets...

Which is approximately 447ft.....

Hope this helped.... ?

8 0

3 years ago

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at

yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

$\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)$

Therefore, the work done by the worker in lifting the bucket is given as:

$W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)$

Now, plug in the values given and solve for 'W'. This gives,

$W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J$

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0

3 years ago

if a cyclist is travelling a road due east at 12km/h and a wind is blowing from south-west at 5km/h. find the velocity of the wi

slamgirl [31]

Answer:

The velocity of wind with respect to cyclist is $-15.5 \widehat{i} - 3.5 \widehat{j}$ .

Explanation:

speed of cyclist = 12 km/h east

speed of wind = 5 km/h south west

Write the speeds in the vector form

$\overrightarrow{vc} = 12 \widehat{i}\\\\overrightarrow{vw} = - 5 (cos 45 \widehat{i} + sin 45 \widehat{j})\\\\\overrightarrow{vw} =-3.5 \widehat{i} - 3.5 \widehat{j}$

The velocity of wind with respect to cyclist is

$\overrightarrow{vw} =-3.5 \widehat{i} - 3.5 \widehat{j}\overrightarrow{v_(w/c)} = \overrightarrow{vw}-\overrightarrow{vc}\\\\\overrightarrow{v_(w/c)} = - 3.5 \widehat{i} - 3.5 \widehat{j} - 12 \widehat{i}\\\\\overrightarrow{v_(w/c)} =-15.5 \widehat{i} - 3.5 \widehat{j}$

7 0

3 years ago

1. Which of these is a relative adverb? where lovely who

PtichkaEL [24]

Who is the right answer

3 0

3 years ago

Read 2 more answers

To determine the amplitude of a wave, you would need to know the . A) distance from the crest to the trough. B) maximum displace

Alexxx [7]

To determine the amplitude of a wave, you would need to know the : B. Maximum displacement of the wave from the equilibrium position

You can find out the displacement by counting the waves that is
written on the Graph

Hope this helps

4 0

3 years ago

Read 2 more answers