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2 years ago

14

A pizza delivery person drives 3 miles North then 5 miles West and then 3 miles South to deliver a pizza. What is the person's d

isplacement?
A. 45 miles
B. 8 miles
C. 11 miles
D. 5 miles west

2 answers:

Anton [14]2 years ago

7 0

Answer: it’s 11

Explanation:

I just did the question

AleksAgata [21]2 years ago

4 0

Answer:11

Explanation:

Just did it

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Which of the following are units that can be used to describe vector

rewona [7]

The answer is D because it’s going by the miles

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3 years ago

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A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s$

$\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}$

$\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}$

$\mu_{k} = 0.704$

b) The speed of the block is determined by using the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = x\cdot \sqrt{\frac{k}{m} }$

$v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }$

$v \approx 9.829\,\frac{m}{s}$

The radius of the circular loop is:

$\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}$

$\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}$

$R = 0.312\,m$

5 0

3 years ago

A 2 kg rock is at the edge of a cliff 20 meters above a lake The rock becomes loose and falls toward the water below. Calculate

natima [27]

Answer:

The potential energy (P.E) at the top is 392 J

The kinetic energy (K.E) at the top is 0 J

The potential energy (P.E) at the halfway point is 196 J.

The kinetic energy (K.E) at the halfway point is 196 J.

Explanation:

Given;

mass of the rock, m = 2 kg

height of the cliff, h = 20 m

speed of the rock at the halfway point, v = 14 m/s

The potential energy (P.E) and kinetic energy (K.E) when its at the top;

P.E = mgh

P.E = (2)(9.8)(20)

P.E= 392 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the top of the cliff = 0

K.E = ¹/₂(2)(0)²

K.E = 0

The potential energy (P.E) and kinetic energy (K.E) at the halfway point;

P.E = mg(¹/₂h)

P.E = (2)(9.8)(¹/₂ x 20)

P.E = 196 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the halfway point = 14 m/s

K.E = ¹/₂(2)(14)²

K.E = 196 J.

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2 years ago

Q1:A large tank is filled with water. The pressure on the base of the fish tank is 4000N/m². The base of the tank is a rectangle

Ymorist [56]

Answer:

F = 36 kN

Explanation:

It is given that,

The pressure on the base of the fish tank is 4000N/m².

The base of the tank is a rectangle measuring 2.0m by 4.5m.

Area of the base of the tank is 9 m²

We need to find the force on the base caused by the base of the water. Pressure on the base of the tank is given by the force acting per unit area such that,

$P=\dfrac{F}{A}\\\\F=P\times A\\\\F=4000\times 9\\\\F=36000\ N\\\\F=36\ kN$

So, the force of 36 kN is acting on the base by the base of water.

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3 years ago

Please help me with these two multiple choice questions!

mafiozo [28]

Answer:

11 is A and 12 is C

Explanation:

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2 years ago