How many meters do you cover in a 10 km (10-K) race?

Trên bóng đèn có ghi là 12v -6w. Khi đèn sáng bình thường thì dòng điện chạy qua đèn có cường độ là

Annette [7]

Answer:

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Explanation:

4 0

2 years ago

Why do you think it is important that the U.S. Constitution defines citizenship?

velikii [3]

Answer:

It's important that the U.S. Constitution defines citizenship because it guarantees it to people who are born within the United States.

Explanation:

It"subjects to the jurisdiction thereof".

5 0

2 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

2 years ago

Help me help me help me

spin [16.1K]

I think d because they are both increasing

7 0

2 years ago

Read 2 more answers

A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one

alekssr [168]

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

Resistance of copper wire=R= $13\Omega$

Resistance is directly proportional to length.

If a wire has greater resistance then,the wire will be greater in length.

Therefore,resistance of long piece of wire=7r

Total resistance of copper wire=Sum of resistance of two piece of wires

$r+7r=13$

$8r=13$

$r=\frac{13}{8}$ ohm

Resistance of long piece of wire= $7\times\frac{13}{8}=\frac{91}{8}\Omega$

Resistance of short piece of wire = $\frac{13}{8}\Omega$

Resistivity of wire and cross section area of wire remains same .

Let L be the total length of wire and L' be the length of short piece of wire.

We know that

$R=\frac{\rho L}{A}$ = $\frac{\rho}{A}L=KL$

$\frac{R}{L}=K$

Where K= $\frac{\rho}{A}$ =Constant

Using the formula

$\frac{13}{L}=\frac{\frac{13}{8}}{L'}$

$\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}$

$L'=\frac{L}{8}$

Length of short piece of wire=L'= $\frac{L}{8}$

Length of long piece of wire= $L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L$

% of length of short piece of wire= $\frac{\frac{L}{8}}{L}\times 100=12.5%$ %

The resistance of the short piece= $\frac{13}{8}\Omega$

The resistance of the long piece= $\frac{91}{8}\Omega$

8 0

3 years ago