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
exis [7]
4 years ago
6

Your friends sit in a sled in the snow. If you apply a force pf 75 N to them, they have an acceleration of 0.9 m/s ^ 2. What is

their mass?
Physics
1 answer:
Elza [17]4 years ago
3 0

Answer:

Mass of the sled in the snow 83.33 kg.

<u>Explanation</u>:

Given that,  

Force applied to move the sled in the snow (F) = 75N

\text { Acceleration }(a)=0.9 \mathrm{m} / \mathrm{s}^{2}

We know that

Newton's second law of motion is  

\text { Force }=\text { mass } \times \text { acceleration }

F = ma (Or "force" is equal to "mass" times "acceleration".)

So if we move this around we can isolate mass and get mass

\text { Mass }=\frac{\text { force }}{\text { accelearation }}

\mathrm{M}=\frac{75}{0.9}

M = 83.33 kg

Mass of the sled in the snow <u>83.33 kg.</u>

You might be interested in
Nuclear sizes are expressed in a unit named<br><br> Magnetism at the centre of a bar magnet is
Illusion [34]
Nuclear size are measured in a unit called fermi
Magnetism at the centre of a bar magnet is zero, because there is no magnetism at all in the centre
Hope this helps :) 
3 0
4 years ago
Read 2 more answers
A kangaroo jumps up with an initial velocity of 36 feet persecond from the ground (assume its starting height is 0 feet).Use the
kkurt [141]

Given

Initial velocity:

36 ft/s

Initial height:

0 ft

Vertical motion model:

h(t) = -16t^2 + ut + s

v = initial velocity

s = is the height

Procedure

We are going to use the model provided for the vertical motion.

\begin{gathered} h(t)=-16t^2+36t+0 \\ h(t)=-16t^2+36t \end{gathered}

We know that at the maximum height the final velocity is 0.

Then we will use the following expression to calculate the maximum height:

\begin{gathered} v^2_f=v^2_o-2ah_{\max } \\ 0=v^2_o-2ah_{\max } \\ 2ah_{\max }=v^2_o \\ h_{\max }=\frac{v^2_o}{2a} \\ h_{\max }=\frac{(36ft/s)^2}{2\cdot32ft/s^2} \\ h_{\max }=20.25\text{ ft} \end{gathered}

Now for time:

\begin{gathered} 20.25=-16t^2+36t \\ 16t^2-36t+20.25=0 \end{gathered}

Solving for t,

\begin{gathered} t_1=2.25 \\ t_2=0 \end{gathered}

The total time the kangaroo takes in the air is 2.3s.

3 0
1 year ago
If2.0J of work is done in raising a 180g apple how far is it lifted?
Alex
We know, W = F * s
W = mg * s

Here, w = 2 J
m = 180 g = 0.180 Kg
g = 9.8 m/s

Substitute their values into the expression:
2 = 0.180*9.8 * s
1.764s = 2
s = 2 / 1.764
s = 1.13 meter

In short, Your Final Answer is 1.13 m

Hope this helps!

7 0
3 years ago
When a carpet is beaten with a stick dust comes out of it ...why
Mandarinka [93]

This can be explain through Newton’s first law of motion where an object at rest remain in rest unless acted upon by an unbalanced force. Hence, the dust particles having inertia and trapped within the pores of the carpet has the tendency to remain at the state of rest, it does not move and resist motion but when the carpet is beaten with a stick, dust comes out because the force from a stick acted upon it.

 

3 0
4 years ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
4 years ago
Other questions:
  • If two wires run parallel and the current passes through both wires in the same direction, which happens to the wires?
    9·1 answer
  • Steve and Carl are driving from Scranton to Bridgeport a distance of 180 miles if they're speed averages 60 miles an hour how lo
    13·1 answer
  • A string is wound tightly around a fixed pulley having a radius of 5.0 cm. as the string is pulled, the pulley rotates without a
    12·2 answers
  • Secondary evidence is the basis for drawing scientific conclusions. <br> a. True<br> b. False
    15·2 answers
  • List the compositional layers in order of most dense to least dense
    12·1 answer
  • What happens when you turned off the gravity on the simulation?
    14·1 answer
  • Define A Gamma Ray .....---_---__☘️​
    9·1 answer
  • What is the effect of gravity on a falling object?
    7·1 answer
  • Use F = 1/T as your basis:
    8·1 answer
  • Suppose a 115 N force is applied to a lawnmower handle at an angle of 25° with the ground and
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!