A cheetah accelerates from rest to 30m/s in 3 seconds. Calculate the acceleration of the cheetah.

A student pushes a 2.85 kg cart causing it to accelerate at a rate of 4.9 m/s squared .What amount of force must the student hav

uysha [10]

In order to find the force (F), you would have to use the formula for it:
F=ma
where m is mass and a is acceleration.
In the problem, the mass is 2.85kg and the acceleration is 4.9m/s^2.
Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.

7 0

3 years ago

8. While taking a measurement, Ajay put the 2nd mark of the scale to the edge of the line and the mark that pointed to the end o

Leni [432]

Answer:

The length of line is 78 cm or 0.78 m.

Explanation:

initial reading 2 mark

final reading 80 cm

The length of the line

= final reading - initial reading

= 80 - 2

= 78 cm

1 cm = 0.01 m

So, 78 cm = 0.78 m

4 0

2 years ago

The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0

katen-ka-za [31]

Answer:

a) F = 2.66 10⁴ N, b) h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

P₂ = P₀ = 1.01 105 Pa

inside

P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

P₁ = 0.85 1.01 10⁵ + 1000 9.8 5

P₁ = 85850 + 49000

P₁ = 1.3485 10⁵ Pa

the net force is

ΔP = P₁- P₂

Δp = 1.3485 10⁵ - 1.01 10⁵

ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

P = Fe / A

F = P A

the area of a circle is

A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

d = 100 cm (1 m / 100 cm) = 1 m

F = 3.385 104 pi / 4 (1) ²

F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

P₁ = P₂

0.85 P₀ + ρ g h = P₀

h = $\frac{P_o ( 1-0.850)}{\rho \ g}$

h = $\frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}$

h = 1.55 m

4 0

2 years ago

Can you help me with this paper please I will give you 20 points!

sweet [91]

1) they are attracting because if you look at the arrows they’re all pointing the same way.

2) if the magnet was turned around they would do the opposite and not attract ( this is called repulsion)

3) magnetic pole

4)magnet

5) magnetic force

6) magnetism

Hope this helps

6 0

2 years ago

Read 2 more answers

Enter an expression for the force constant for the floating raft, in terms of L, g, and the density of water, ρ.

andriy [413]

Answer:

K = ρL²g

Explanation:

Consider L as the length of the raft inside the water when the raft is displaced through additional distance y;

Then:

F = upthrust ( restoring force) = weight of the liquid displaced.

$F = V_{\omega} \rho_{\omega} g= A y \rho_{\omega} g$