1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Lemur [1.5K]
3 years ago
8

If Angela ran to the bus and back to where she started what distance would she travel ?

Physics
1 answer:
Yakvenalex [24]3 years ago
5 0

Answer:

Distance = 30 m

Explanation:

Since the position of the bus is at zero, and distance is a scalar quantity. That is, we are only concerned about the magnitude

If Angela ran to the bus and back to where she started, the distance travelled will be:

Distance = 15 + 15 = 30 m

But her displacement will be:

Displacement = 15 - 15 = 0.

You might be interested in
Did you know that your answer automatically gets reported if you have "idk" anywhere in the answer?
Maurinko [17]

Answer:

rlly??

Explanation:

4 0
3 years ago
Read 2 more answers
How much heat is needed to warm 0.072kg of gold from 20 celsius and 90 celsius if the specific heat of gold 136 joules
dybincka [34]

Heat supplied to the gold will raise the temperature of the gold from 20 degree Celsius to 90 degree Celsius.

Mass of the gold (m) = 0.072 kg

Temperature change (ΔT) = 90 - 20 = 70 degree Celsius

Specific heat capacity of the gold (c) = 136 J/kg C

Heat supplied = m × c × ΔT

Heat supplied = 0.072 × 136 × 70

Heat supplied = 685.44 Joules

Hence, the heat supplied to the gold to raise the temperature from 20 degree Celsius to 90 degree Celsius = 685.44 Joules

5 0
3 years ago
A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. If your push was directed at an angle
emmasim [6.3K]

<u>Given data:</u>

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ       --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ  =>  R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>

<em>Pmin = W sin( α +Θ  )</em>

<em>          = W[ sin α.Cos Θ + cos α.sin Θ]</em>

<em>           = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>

<em>           = 460.9 N</em>

<em>Minimum push required to move the box up the ramp is 460.9 N</em>


7 0
3 years ago
Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

5 0
3 years ago
Calculate the mass of a liquid with a density of 2.5 g/ml and a volume of 15ml
Nana76 [90]
Using the density equation and clearing mass:

\rho = \frac{m}{V}\ \to\ m = \rho\cdot V = 2.5\frac{g}{mL}\cdot 15\ mL = \bf 37.5\ g
3 0
3 years ago
Other questions:
  • What current flow (I) is associated with an input voltage of 5.0V and resistors R1 = 1.5 kiloohms and R2 = 0.5 kiloohms? Calcula
    11·1 answer
  • What factors affect the speed of a wave? Check all that apply.
    8·2 answers
  • I’m just completely confused
    15·1 answer
  • 1 Do alien exist<br>2 why sky is blue<br>3 what is the weight of sky<br>4 why occasion is blue​
    12·2 answers
  • A roller coaster has a mass of 650 kg. It sits at the top of a hill with height 78 m. If it drops from this hill, how fast is it
    13·1 answer
  • A long wire carrying I=1.1A of constant current oriented North-to-South (and the current is running northwards) is placed horizo
    12·1 answer
  • The composition and properties of a(n) _____ are always constant throughout a given sample and from sample to sample
    13·1 answer
  • How long does it take light ray B to reach the girls eye? (speed of light at 3*10^8) And could someone explain why?
    10·1 answer
  • A student walks 3 north and 4 m west. The magnitude of the resultant displacement for the student is
    6·1 answer
  • Which fact is true during a fission chain reaction?
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!