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3 years ago

50 points for this please help

Physics

2 answers:

Softa [21]3 years ago

7 0

but the points are only 25

wel3 years ago

6 0

Answer:

are they talking about the form classification or..?

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If object A has more mass than object B, what will object A need to accelerate at the same rate as object B?

Leni [432]

Answer:

More force

Explanation:

Object A has more mass than object B

For object A to accelerate at the same rate as object B, it will need more force.

According to Newton's second law of motion "the net force on a body is the product of its mass and acceleration".

Net force = mass x acceleration

Now, if a body has more mass and needs to accelerate at the same rate as another one with a lower mass, the force on it must be increased.

3 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

An airplane weighing 11,000 N climbs to a

Gennadij [26K]

The power in horsepower is 40.1 hp

Explanation:

We start by calculating the work done by the airplane during the climb, which is equal to its change in gravitational potential energy:

$W=(mg)\Delta h$

where

mg = 11,000 N is the weight of the airplane

$\Delta h = 1.6 km = 1600 m$ is the change in height

Substituting,

$W=(11,000)(1600)=17.6\cdot 10^6 J$

Now we can calculate the power delivered, which is given by

$P=\frac{W}{t}$

where

$W=17.6\cdot 10^6 J$ is the work done

$t=9.8 min \cdot 60 = 588 s$ is the time taken

Substituting,

$P=\frac{17.6\cdot 10^6 J}{588}=2.99\cdot 10^4 W$

Finally, we can convert the power into horsepower (hp), keeping in mind that

$1 hp = 746 W$

Therefore,

$P=\frac{2.99\cdot 10^4}{746}=40.1 hp$

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0

3 years ago

Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay

kaheart [24]

Answer:

0.278 m/s

Explanation:

We can answer the problem by using the law of conservation of momentum. In fact, the total momentum before the collision must be equal to the total momentum after the collision.

So we can write:

$mu=(m+M)v$

where

m = 0.200 kg is the mass of the koala bear

u = 0.750 m/s is the initial velocity of the koala bear

M = 0.350 kg is the mass of the other clay model

v is their final combined velocity

Solving the equation for v, we get

$v=\frac{mu}{m+M}=\frac{(0.200)(0.750)}{0.200+0.350}=0.278 m/s$

8 0

3 years ago

If a 5 N force pushes a 20 kg mass on a frictionless surface, how fast is the mass going in 5 seconds.

stira [4]

Explanation:

Let me know if you have questions

3 0

3 years ago