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3 years ago

Please help me ,create a slogan that promotes the role of internet in education

Physics

1 answer:

aleksklad [387]3 years ago

7 0

Answer:

¨Facts you didn´t know¨ or ¨unknown facts¨

Explanation:

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When are all the forces acting upon an object balanced?

Rudik [331]

Answer:

When two forces acting on an object are equal in size but act in opposite directions, we say that they are balanced forces.

Explanation:

7 0

2 years ago

Momentum is a vector quantity that depends

MissTica

On the change in potential energy

3 0

3 years ago

True or False: As distance between 2 objects increases, the gravitational force decreases

VikaD [51]

Answer:

True

Explanation:

Because it is:)

5 0

3 years ago

A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential

castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge $q=+10.5\times10^{-6}\ C$

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot d$

$\Delta U=-q(E\cdot d)$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}$

$\Delta U=-3.32\times10^{-3}\ J$

The change in electric potential energy is $-3.32\times10^{-3}\ J$

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot (-d)$

$\Delta U=-q(E\cdot (-d))$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})$

$\Delta U=3.32\times10^{-3}\ J$

The change in electric potential energy is $3.32\times10^{-3}\ J$

Hence, This is the required solution.

3 0

3 years ago

The main purpose of an air bag is to stop a passenger during a car accident in a greater amount of time than if the air bag were

Simora [160]

Answer:

a) 45571 N

b) 22786 N

c) 4557 N

Explanation:

Since the goal of the airbag is helping the person to stop after the collision in a greater time, this means that the change in momentum must finish when this is just zero.
In other words, the change in momentum, must be equal to the initial one, but with opposite sign.

$\Delta p = - p_{o} = -m*v = -55 kg*29m/s = -1595 kgm/s (1)$

Now, just applying the original form of Newton's 2nd Law, we know that this change in momentum must be equal to the impulse needed to stop the person:

$\Delta p = F* \Delta t (2)$

So, as we know the magnitude of Δp from (1) and we have different Δt as givens, we can get the different values of F (in magnitude) required to stop the person for each one of them, as follows:

$F_{1} = \frac{\Delta p}{\Delta t_{1}} = \frac{1595kgm/s}{0.035s} = 45571 N (3)$

$F_{2} = \frac{\Delta p}{\Delta t_{2}} = \frac{1595kgm/s}{0.07s} = 22786 N (4)$

$F_{3} = \frac{\Delta p}{\Delta t_{3}} = \frac{1595kgm/s}{0.35s} = 4557 N (5)$

4 0

3 years ago