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3 years ago

Which of the following are likely to be affected by a magnet? Check all that apply.

Physics

2 answers:

Gelneren [198K]3 years ago

6 0

<h3><u>Answer;</u></h3>

C.An access card you scan for entry into a building

<h3><u>Explanation;</u></h3>

<em><u>Magnets are materials that have the capability of producing magnetic field.</u></em> Magnets attract magnetic material such as iron and cobalt while they don't have an effect on non-magnetic materials such as wood and plastic.
<em><u>Access cards are plastic cards with a microchip or a magnetic strip containing an encoded data that is read for an individual to gain entry into a building.</u></em>
<em><u>Due to the presence of a magnetic strip that is swiped through the mechanism in the door, they may be affected by magnetic field from the magnets.</u></em>

stellarik [79]3 years ago

5 0

I think the correct answers from the choices listed above are options B and C. A credit card and an access card you scan for entry into a building are most likely to be affected by a magnet. Hope this answers the question. Have a nice day.

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An insulated thermos contains 106.0 cm3 of hot coffee at a temperature of 80.0 °C. You put in 11.0 g of ice cube at its melting

Andrei [34K]

Answer:

the final temperature is T f = 64.977 ° C≈ 65°C

Explanation:

Since the thermus is insulated, the heat absorbed by the ice is the heat released by the coffee. Thus:

Q coffee + Q ice = Q surroundings =0 (insulated)

We also know that the ice at its melting point , that is 0 °C ( assuming that the thermus is at atmospheric pressure= 1 atm , and has an insignificant amount of impurities ).

The heat released by coffee is sensible heat : Q = m * c * (T final - T initial)

The heat absorbed by ice is latent heat and sensible heat : Q = m * L + m * c * (T final - T initial)

therefore

m co * c co * (T fco - T ico) + m ice * L + m ice * c wat * (T fwa - T iwa) = 0

assuming specific heat capacity of coffee is approximately the one of water c co = c wa = 4.186 J/g°C and the density of coffee is the same as water

d co = dw = 1 gr/cm³

therefore m co = d co * V co = 1 gr / cm³ * 106 cm³ = 106 gr

m co * c wat * (T f - T ico) + m ice * L + m ice * c wat * (T f - T iwa) = 0

m co * c wat * T f+ m ice * c wat * T f = m ice * c wat * T iwa + m co * c wat * Tico -m ice * L

T f = (m ice * c wat * T iwa + m co * c wat * Tico -m ice * L ) /( m co * c wat * + m ice * c wat )

replacing values

T f = (11 g * 4.186 J/g°C * 0°C + 106 g * 4.186 J/g°C*80°C - 11 g * 334 J/gr) / ( 11 g * 4.186 J/g°C + 106 g * 4.186 J/g°C* ) = 64,977 ° C

T f = 64.977 ° C

7 0

3 years ago

A 1000 kg race car rounds a curve with a radius of 50 m.

Kruka [31]

Answer:

5644556677888777766554433

5 0

2 years ago

1. Define friction. Name and describe two kinds of friction.

jeka57 [31]

Answer:

Friction is a force that holds back the movement of a sliding object.

Explanation:

The two types of friction: Static friction and Kinetic friction. Static friction operates between two surfaces that aren't moving relative to each other, while kinetic friction acts between objects in motion.

6 0

3 years ago

A bus is traveling with a uniform acceleration of 2.75 meters/second2. If the initial velocity of the bus is 16.5 meters/second,

Kaylis [27]

<span>The velocity will be 41.25 m/s2 after 9 seconds. To find velocity after a specific time period, multiply the acceleration (2.75) times the number of seconds (9) to receive 24.75 m/s, then add that to the initial velocity of 16.5 m/s. 24.75 + 16.5 = 41.25 m/s2.</span>

8 0

3 years ago

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$

the vertical component of the velocity is

$v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.$

which gives a speed $v$ of

$v = \sqrt{v_x^2+v_y^2}$

$\boxed{v =33.3m/s.}$

4 0

2 years ago