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MAXImum [283]
3 years ago
10

An object accelerating at 16 m/s/s doubles its mass and triples its net force acting on it. What will the new acceleration be? (

Just number no units)
Physics
1 answer:
nataly862011 [7]3 years ago
6 0

Answer:

24 m/s²

Explanation:

The given parameters are;

The initial acceleration of the object, a = 16 m/s²

Let 'm' represent the initial mass of the object

The initial force acting on the object, F = m × a

∴ F = 16 × m = 16·m

When the mass is doubled, we have;

The new mass of the object, m₂ = 2 × m = 2·m

When the net force acting on the object triples, we have;

The new net force acting on the object, F₂ = 3 × F = 3 × 16·m = 48·m

From F = m × a, we have;

a = F/m

∴ The new acceleration of the object, a₂ = F₂/m₂

From which, by plugging in the values, we have;

a₂ = 48·m/(2·m) = 24

The new acceleration of the object, a₂ = 24 m/s².

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A ball is swung in a horizontal circle at a constant speed. Each circle takes 0.85 seconds to complete and the rope is 0.40 m lo
BaLLatris [955]

Answer:

The centripetal acceleration will be "21.785 m/s²".

Explanation:

The given values are:

Time,

t = 0.85 seconds

Length of rope,

r = 0.40 m

Mass of ball,

m = 0.80 kg

As we know,

⇒ w=\frac{2 \pi}{t}

On substituting the values, we get

⇒      =\frac{2\times 3.14}{0.85}

⇒      = \frac{6.28}{0.85}

⇒      =7.38 \ rad/s^2

The centripetal acceleration will be:

⇒  a=r\times w^2

⇒     =0.40\times (7.38)^2

⇒     =0.40\times 54.46

⇒     =21.785 \ m/s^2

7 0
2 years ago
Whats the Independent and Dependent Variables
astra-53 [7]
The independent variable is the type of fuel used and the dependent variable is the speed of the race car. The independent variable could be changed through the experimental process to see its relation with the dependent variable<span>. The dependent variable is the result of the independent variable changes.</span>
5 0
3 years ago
A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
Juan compró un carro que dicen que es muy rápido. Cuándo lo probó recorrió una distancia de 4500m en tan solo 5 min. ¿Qué veloci
sweet [91]

Answer:

Don't know sorry..................

4 0
3 years ago
HELPPP
AlekseyPX
Friction is caused by the uneven surfaces of touching objects
3 0
2 years ago
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