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Cloud [144]
3 years ago
13

A 2-Newton upward net force is being applied to a 10-kg object. What is the magnitude of the upward

Physics
1 answer:
Colt1911 [192]3 years ago
7 0
Acceleration in m/s^2 = 2/10 = 0.2 m/s^2
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The alternating current which crosses an apparatus of 600 W has a maximum value of 2.5 A. What is efficient voltage between its
azamat

Answer: Option (b) is correct.

Explanation:

Since we know that,

P = VI

where;

P = power

V= Voltage

I = Current

Since it's given that,

P = 600W

I = 2.5 A

equating these values in the above equation, we get;

<em>V = \frac{600}{2.5}</em>

<em>V = 240 V</em>

8 0
3 years ago
Which of two factors influence the weight of an object due to gravitational pull?
densk [106]
The weight of an object when calculated by multiplying with the pull of the gravity is dependent on the mass of the object and the value of g. The value of g is constant however is still dependent on the distance of the object from the center of the Earth. Thus, the answers are <em>mass and distance. </em>
5 0
3 years ago
Which of the following examples does not make use of total internal reflection?
liubo4ka [24]

Answer:

laser pointer does not make use of total internal reflection

7 0
3 years ago
Friends Burt and Ernie stand at opposite ends of a uniform log that is floating in a lake. The log is 3.0 m long and has mass 20
Taya2010 [7]

Answer:

The distance the log has moved by the time Ernie reaches Bur is 1.33 m.

Explanation:

give information:

The log is 3.0 m long and has mass 20.0 kg.

Burt has mass 30.0 kg; Ernie has mass 40.0 kg

Ernie has mass 40.0 kg.

to find the distance, first, we have to calculate the center of mass

X = ∑ m x /∑m

   = (20 x (3/2)) + (30 x 0) + (40 x 3)/ (20+30+40)

   = 150/90

   = 5/3

when Ernie walk, the center of the mass is

X = (70 x 0) + (20 x (3/2))/(70 + 20)

  = 30/90

  = 1/3

the distance of log moved = 5/3 - 4/3 = 1.33 m

5 0
3 years ago
astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating o
devlian [24]

Answer:

The right answer is:

(a) 63.83 kg

(b) 0.725 m/s

Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

The mass of the string will be:

⇒ T=2 \pi\sqrt{\frac{m}{k} }

or,

⇒ m=\frac{k T^2}{4 \pi^2}

On substituting the values, we get

⇒     =\frac{280\times (3)^2}{4 \pi^2}

⇒     =\frac{280\times 9}{4\times (3.14)^2}

⇒     =68.83 \ kg

(b)

The speed of the string will be:

⇒  \frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2

then,

⇒             v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }

On substituting the values, we get

⇒                =\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }

⇒                =\sqrt{\frac{280(0.16-0.04)}{63.83} }

⇒                =\sqrt{\frac{280\times 0.12}{63.83} }

⇒                =0.725 \ m/s

4 0
3 years ago
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