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S_A_V [24]
2 years ago
11

Which statement best describes the acceleration of a baseball?

Physics
1 answer:
insens350 [35]2 years ago
3 0

Answer:

C the baseball began at rest and rolls at a rate of 14.7 m/s after 1.5 seconds

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A force gives a 5.0 kg object an acceleration of 2.0 m/s 2. The same force would give a 20 kg object an acceleration of _____. 0
Oksi-84 [34.3K]

m = 5 kg

a = 2 m/s²

to find the force that accelerates the 4 kg object @ 2 m/s²

F = ma = 5 kg x 2 m/s² = 10 N

To find what acceleration 10 N would give a 20 kg object

a = F/m = 10 N/20 kg = 0.5 m/s

6 0
2 years ago
Read 2 more answers
The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a
dimaraw [331]

The absolute pressure is given by the equation,

P_{abs}=P_{atm}-P_{vac}

Here, P_{abs} is absolute pressure,P_{atm} is atmospheric pressure and P_{vac} is vacuum pressure.

Therefore,

P_{abs}=98 kPa-80 kPa=18kPa

The gage pressure is given by the equation,

P_{gage}=P_{abs}-P_{atm}.

Thus,

P_{gage}=18kPa-98 kPa=-80 kPa.

In kn/m^2,

The absolute pressure,

P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2

The gage pressure,

P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2.

In lbf/in2

The absolute pressure,

P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2

The gage pressure,

P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2

In psi,

The absolute pressure,

P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi.

The gage pressure,

P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi

In  mm Hg

The absolute pressure,

P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg

The gage pressure,

P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg



3 0
2 years ago
This phenomenon is known as the bends. If a scuba diver rises quickly from a depth of 25.0 m in Lake Michigan (which is fresh wa
Sveta_85 [38]

Explanation:

Attachments of the solution are below.

6 0
3 years ago
A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is
Aleksandr [31]

Answer:

The acceleration of the crate is 1.8 m/s² so the answer is a.

Explanation:

The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)

So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.

As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:

Starting with the sum of forces in the y-direction, we get:

ΣF_{y}=0

We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.

so the sum will be:

N-W=0

when solving for N we get that:

N=W

where W is found by multiplying the mass of the crate by the acceleration of gravity:

N=250kg*9.8m/s²

N=2450N

Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:

f_{k}=Nμ

where μ is the kinetic friction coefficient.

So we get that the kinetic friction is:

f_{k}=2450N*0.12

so

f_{k}=294

With this information we can go ahead and find the sum of horizontal forces:

ΣF_{x}=ma

In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.

so the sum of forces look like this:

750N-f_{k}=ma

so

750N-294N=(250kg)a

when solving for a we get:

a=\frac{759N-294N}{250kg}\\ \\a=1.8m/s^{2}

so the crate's acceleration is 1.82m/s².

5 0
3 years ago
Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at
Wewaii [24]

Answer:

θ=19.877⁰

Explanation:

Given data

Velocity Va=34.0 km/h

Velocity Va=100 km/h

To find

Angle θ

Solution  

We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got

Sinθ=(Va/Vb)

Sinθ=(34.0/100)

θ=Sin⁻¹(34.0/100)

θ=19.877⁰

5 0
3 years ago
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