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ch4aika [34]
3 years ago
10

A squirrel is trying to locate some nuts he buried for the winter. He moves 4.3 m to the right of a stone and dogs unsuccessfull

y. Then he moves 1.1 m to the left of his hole, changes his mind, and moves 6.3 m to the right of that position and digs a second hole. No luck. Then he moves 8.0 m to the left and digs again. He finds a nut at last. What is the squirrel's total displacement from his starting point
Physics
1 answer:
krek1111 [17]3 years ago
8 0

Answer:

The total displacement from the starting point is 1.5 m.

Explanation:

You need to sum and substract, depending on the movement (to the right, sum; to the left, substract).

First, it moves 4.3 m right and return 1.1 m. So the new distance from the starting point is 3.2 m.

Second, it moves 6.3 m right, so the new distance is 9.5 m.

Finally it moves 8 m to the left, so 9.5 m - 8 m= 1.5 m.

Summarizing, at the end the squirrel is 1.5 m from its starting point.

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Calculate the pressure exerted by a 4000N camel on the sand. The camel’s feet have a
Harrizon [31]

Answer:

The pressure exerted by camel feet is <u>2000 N/m²</u>.

Step-by-step explanation:

<h3><u>Solution</u> :</h3>

Here, we have given that ;

  • Force applied on camel feet = 4000 N
  • Total area of camel feet = 2 m²

We need to find the pressure exerted by camel feet.

As we know that :

{\longrightarrow{\pmb{\sf{Pressure= \dfrac{Area}{Force}}}}}

Substituting all the given values in the formula to find the pressure exerted by camel feet.

\begin{gathered} \begin{array}{l} {\longrightarrow{\sf{Pressure= \dfrac{Area}{Force}}}} \\  \\ {\longrightarrow{\sf{Pressure= \dfrac{4000}{2}}}}  \\  \\ {\longrightarrow{\sf{Pressure= \cancel{\dfrac{4000}{2}}}}} \\  \\ {\longrightarrow{\sf{Pressure= 2000 \: N/{m}^{2}}}} \\  \\\star \:  \small\underline{\boxed{\sf{\purple{Pressure= 2000 \: N/{m}^{2}}}}} \end{array}\end{gathered}

Hence, the pressure exerted by camel feet is 2000 N/m².

\rule{300}{2.5}

3 0
2 years ago
Light in the air is incident at an angle to the surface of (12.0 A) degrees on a piece of glass with an index of refraction of (
Orlov [11]

The question is incomplete. You dis not provide values for A and B. Here is the complete question

Light in the air is incident at an angle to a surface of (12.0 + A) degrees on a piece of glass with an index of refraction of (1.10 + (B/100)). What is the angle between the surface and the light ray once in the glass? Give your answer in degrees and rounded to three significant figures.

A = 12

B = 18

Answer:

18.5⁰

Explanation:

Angle of incidence i = 12.0 + A

A = 12

= 12.0 + 12

= 14

Refractive index u = 1.10 + B/100

= 1.10 + 18/100

= 1.10 + 0.18

= 1.28

We then find the angle of refraction index u

u = sine i / sin r

u = sine24/sinr

1.28 = sine 24 / sine r

1.28Sine r = sin24

1.28 sine r = 0.4067

Sine r = 0.4067/1.28

r = sine^-1(0.317)

r = 18.481

= 18.5⁰

4 0
2 years ago
A golf club with mass 0.5 kg is swung at 23 m/s toward a stationary golf ball on a tee. The club barely slows down to 17 m/s, bu
mylen [45]

Answer:

I gotchu fam. 0.0435 kg

Explanation:

m1v1+m2v2=m1v1f+m2v2f

(0.5)(23)+(m2)(0)=(0.5)(17)+(m2)(69)

11.5=8.5+69m2

3=69m2

0.0435 kg

5 0
2 years ago
Which of the following changes will increase the period of an oscillating spring mass system?
Gnoma [55]

Answer:

a. an increase in the mass on the spring.

Explanation:

T = 2π/ω = 2π/√(k/m) = 2π√(m/k)

2π is a constant

as m is in the numerator, increasing mass will increase the period.

6 0
3 years ago
When a ball rolls down a hill, what energy conversion occurs?
Ludmilka [50]

Answer:

C

Explanation:

When a ball rolls down a hill, Potential Energy Conversion takes place to Kinetic Energy.

8 0
3 years ago
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