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3 years ago

Define the term “force”.

Physics

2 answers:

romanna [79]3 years ago

6 0

The Jedi have the force. it flows through all things. Joking it will change the motion of an object

Aleonysh [2.5K]3 years ago

3 0

Energy that is applied to an object.

--TheOneandOnly003

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The primary coil of an ideal transformer has 100 turns and its secondary coil has 400 turns. if the ac voltage applied to the pr

Nana76 [90]

The formula used in calculations relating to transformers is:

<span>Secondary voltage (Vs)/ Primary voItage (VP) = Secondary turns (nS)/ Primary turns (nP)</span>

Substituting the given values to find for Vs,

Vs / 120 V = 400 turns / 100 turns

6 0

4 years ago

Anna35 [415]

Answer:

430.

Explanation:

If we know that 0.5 is half of a whole number, then we can simply understand that we need to 215 x 2 to get our answer.

3 0

3 years ago

A man pushed on the side ..

TiliK225 [7]

Answer:

B.will increase the maximum static friction between the box and the floor

Explanation:

Because static friction is the force that keeps an object at rest

7 0

3 years ago

A truck with a mass of 1330 kg and moving with a speed of 15.0 m/s rear-ends a 805 kg car stopped at an intersection. The collis

Dovator [93]

Answer:

The Speed of the vehicles is 9.34m/s

Explanation:

For an elastic collision the two bodies move with similar velocities after collision

Given

M1=1330kg

V1=15m/s

M2=805kg

V2=0(the car is parked on neutral)

The formula is

M1V1+M2V2=(M1+M2)V

1330*15+805*0=(1330+805)V

19950+0=2135V

2135V=19950

divide both sides by 2135

V=19950/2135

V=9.34m/s

8 0

3 years ago

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

c) $v_b = -v cos \omega t$

$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

$Y =- 0.682 \times sin (5.92 \times 0.42)$

$Y =- 0.42$

Force = $m \omega^2 |Y|$

= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

4 0

3 years ago