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3 years ago

In the equation Δp=m(vf−vi), which quantity is considered to be constant?

Physics

1 answer:

SVEN [57.7K]3 years ago

6 0

The m value because it’s mass. This will not change throughout the problem because an object can not gain mass

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A 1200N load is to be lifted with 200N effort using a first class lever. At what distance

kipiarov [429]

Explanation:

Hey there!!

Let's simply work with it.

Here,

load = 1200N

Effort = 200N

Load distance = 15cm

We have,

According to the principle of lever.

L×LD = E×ED.

1200×15 = 200× ED.

18000 = 200ED.

$ed = \frac{18000}{200}$

Therefore, Effort Distance = 90cm.

Hope it helps....

7 0

3 years ago

Suppose you take

kati45 [8]

Answer:

a current will be induced.

Explanation:

4 0

3 years ago

Read 2 more answers

If two children, with masses of 16 kg and 24 kg , sit in seats opposite one another, what is the moment of inertia about the rot

Elena-2011 [213]

Answer:

The moment of inertia about the rotation axis is 117.45 kg-m²

Explanation:

Given that,

Mass of one child = 16 kg

Mass of second child = 24 kg

Suppose a playground toy has two seats, each 6.1 kg, attached to very light rods of length r = 1.5 m.

We need to calculate the moment of inertia

Using formula of moment of inertia

$I=I_{1}+I_{2}$

$I=(m+m_{1})\times r^2+(m+m_{2})\times r^2$

m = mass of seat

m₁ =mass of one child

m₂ = mass of second child

r = radius of rod

Put the value into the formula

$I=(16+6.1)\times(1.5)^2+(24+6.1)\times(1.5)^2$

$I=117.45\ kg-m^2$

Hence, The moment of inertia about the rotation axis is 117.45 kg-m²

8 0

4 years ago

A car enters a freeway with a speed of 6.5 m/s and accelerates to a speed of 24 m/s in 3.5 min. How far does the car travel whil

Vesna [10]

Explanation:

Using Equations of Motion :

(1) v = u + at

24 = 6.5 + a * 210

a (Acceleration) = 0.083 m/s^2

(2) v^2 = u^2 + 2aS

S = 576 - 42.25 / 0.166

S (Distance travelled) = 3215.3 m 

(Option A seems a typo since the answer is 3215.3 m)

4 0

3 years ago

An experimenter using a gas thermometer found the pressure at the triple point of water (0.01°C) to be 4.80 × 10⁴ Pa and the pre

irakobra [83]

Answer:

$T = -282.33^o C$

Explanation:

As we know that the relation between temperature and pressure is a linear relation

so we have

$P - P_o = \frac{P_1 - P_o}{T_1 - T_o} (T - T_o)$

here we know that

$P_1 = 6.50 \times 10^4$

$P_o = 4.80 \times 10^4$

$T_1 = 100^o C$

$T_o = 0.01^o C$

now we will have

$P - 4.80 \times 10^4 = \frac{(6.50 - 4.80)\times 10^4}{100 - 0.01}(T - 0.01)$

$P = 4.80 \times 10^4 + 170.02(T - 0.01)$

now if P = 0

then we will have

$0 = 4.80 \times 10^4 + 170.02(T - 0.01)$

$T = -282.33^o C$

7 0

3 years ago