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Lemur [1.5K]
3 years ago
9

I Require Help on this Physics Question, I made an attempt at answering but I just have no idea.

Physics
1 answer:
Mashutka [201]3 years ago
5 0

Answer:

It traveled 24 centimeters

Explanation:

The displacement of the object is equal to the area under the velocity vs time graph.

We can split this graph into two shapes, a triangle and a rectangle.  So the total area is:

A = ½bh + wh

A = ½ (4 s − 0 s) (4 cm/s) + (8 s − 4 s) (4 cm/s)

A = 8 cm + 16 cm

A = 24 cm

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A butcher grinds 5 and 3/4 lb of meat then sells it for 2 and 2/3 pounds to the customer what is the maximum amount me that the
KiRa [710]

Answer:

The maximum amount of meat that the butcher can sell is  3\frac{1}{12}\:lb

Explanation:

The maximum amount can be found by taking the difference of mixed numbers.

5\frac{3}{4}-2\frac{2}{3}\\\\\mathrm{Subtract\:the\:numbers:}\:5-2=3\\\\\mathrm{Combine\:fractions:\:}\frac{3}{4}-\frac{2}{3}=\frac{1}{12}\\\\=3\frac{1}{12}\\

Best Regards!

8 0
3 years ago
An interstellar space probe is launched from Earth. After a brief period of acceleration it moves with a constant velocity, 70.0
sleet_krkn [62]

Answer:

22.26 years

, 15.585 light years  , 11.13 light years

Explanation:

a)

t' = t/(\sqrt{1-(v/(c*v)/c)}

= 15.9/\sqrt{(1-0.7*0.7)}

= 22.26 years

b)

0.7*c*22.26 years

=15.585 light years  

c)

0.7*c*15.9

=11.13 light years

3 0
3 years ago
A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.
Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck  moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s

Calculation of the acceleration of the  hockey puck

We apply the Formula (1)

vf=v₀+a*t      v₀=5.8 m/s ,  vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t   Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t²   ,  d=15.1 m ,  v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²  

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the  frictional force (Ff)

We apply Newton's second law

∑F = m*a    Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the  hockey puck in the attached graphic

∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²

-Ff = 0.115*(-1.115)  We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck  Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N =  1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0
3 years ago
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The gravitational attraction between two objects will what is the object move further apart
finlep [7]
The farther apart the two objects, the weaker the gravitational attraction between them.
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3 years ago
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