Joe runs 10 m north, 20 m south, 9m south, and then 15 m north. What is Joe's<br> displacement?

A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 3.30 m/s . The coefIficient o

Artemon [7]

Answer:

Explanation:

Mass =11.2kg

Constant velocity =3.3m/s

μk=0.25

Since the body is moving in constant velocity, then the acceleration is zero(0).

ΣF = Σ(ma)

The normal force acting on the body is upward and the weight is acting downward

Then ΣFy=0

Therefore, N=W

W=mg=11.2×9.8=109.76N

So, N=W=109.76N

Frictional force is given as

Fr=μkN

Fr=0.25×109.76

Fr=27.44N

Frictional force acting against the motion is 27.44N

Then the forward force moving the body forward

ΣF = Σ(ma)

Since a = 0

Then,

ΣF = 0

F-Fr=0

Then F=Fr

So the force moving the body forward is 27.44N

8 0

3 years ago

The speed of a wave on a violin A string is 288 m/s and on the G string is 128 m/s. The force exerted on the ends of the string

Katyanochek1 [597]

Answer:

$\dfrac{\mu_A}{\mu_G}=0.197$

Explanation:

given,

Speed of a wave on violin A = 288 m/s

Speed on the G string = 128 m/s

Force at the end of string G = 110 N

Force at the end of string A = 350 N

the ratio of mass per unit length of the strings (A/G). = ?

speed for string A

$v_A = \sqrt{\dfrac{F_A}{\mu_A}}$ .......(1)

speed for string G

$v_G = \sqrt{\dfrac{F_G}{\mu_G}}$ ........(2)

Assuming force is same in both the string

now,

dividing equation (2)/(1)

$\dfrac{v_G}{v_A}=\dfrac{\sqrt{\dfrac{F_G}{\mu_G}}}{\sqrt{\dfrac{F_A}{\mu_A}}}$

$\dfrac{v_G}{v_A}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}$

$\dfrac{128}{288}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}$

$\dfrac{\mu_A}{\mu_G}=0.197$

5 0

2 years ago

Scientists can help us understand the impact of human activities on the environment. True or false?

zhannawk [14.2K]

Answer:

True

Explanation:

hope it will help you

5 0

2 years ago

The radius of Earth is about 6450 km. A 7070 N spacecraft travels away from Earth. What is the weight of the spacecraft at a hei

Triss [41]

Answer:

(a) 1767.43 N

(b) 182.45 N

Explanation:

Radius of earth, R = 6450 km

Weight of person, W = 7070 N

mass of person, m = W / g = 7070 / 9.8 = 721.4 kg

(a) h = 6450 km

The value of acceleration due to gravity on height is given by

$g' = g\left ( \frac{R}{R+h} \right )^2$

$g' = g\left ( \frac{6450}{6450+6450} \right )^2$

g' = g / 4 = 9.8 / 4 = 2.45 m/s^2

The weight of the person at such height is

W' = m x g' = 721.4 x 2.45

W' = 1767.43 N

(b) h = 33700 km

The value of acceleration due to gravity on height is given by

$g' = g\left ( \frac{R}{R+h} \right )^2$

$g' = g\left ( \frac{6450}{6450+33700} \right )^2$

g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2

The weight of the person at such height is

W' = m x g'

W' = 721.4 x 0.253

W' = 182.45 N

3 0

3 years ago

I am a give awayer please like this and say thanks

s344n2d4d5 [400]

Answer:

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4 0

2 years ago

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Joe runs 10 m north, 20 m south, 9m south, and then 15 m north. What is Joe's displacement?