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aleksandr82 [10.1K]
3 years ago
12

The total mass of the wheelbarrow and the road is 80 kg calculate the weight of the wheelbarrow and the road

Physics
1 answer:
Alja [10]3 years ago
7 0

Answer:

The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.

Explanation:

Given that,

The total mass of the wheelbarrow and the road is 80 kg.

The weight of an object is given by :

W = mg

where

g is acceleration due to gravity

So,

W = 80 × 9.8

= 784 N

So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.

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A physics student of mass 51.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend
balandron [24]

Answer:

Explanation:

The wheel and falling student will have common acceleration .

For rotational motion of wheel

Tx r = I α , T is tension in the crank , α is angular acceleration of wheel  , I is moment of inertia , r is radius of the wheel.

= I a / r

T = I a / r²

For motion of student

Mg - T = Ma , M is mass of the wheel.

Mg - I a / r²  = Ma

Mg = Ma +I a / r²

Mg = (M +I  / r²)a

a = Mg / (M +I  / r²)

= 51 x 9.8 / ( 51 + 9.6 / .3² )

499.8 / (51+ 106.67 )

= 499.8 / 157.67

= 3.17 m / s².

If time t is taken to  fall by 12 m

12 = 1/2 a t²

24 / a = t²

24 / 3.17 =t²

t²= 7.57

t = 2.75 s

velocity to reach sidewalk

v = u + at

= 3.17  x 2.75

= 8.72 m / s

5 0
3 years ago
A train travels 2975 miles in 3 days how far is it moving per day?
uysha [10]

Answer:

991.67 miles per day

Explanation:

Since it travels 2975 miles per day, distance traveled in a day =2975/3=991.67 miles

7 0
3 years ago
How many grams of CO-60 result in 1 Millicuire of activity? How many years until the activity decays to 1 microcuire tl/2 =5.3 Y
vredina [299]

Answer:

m =8.81*10^{-6}grams

time t = 52.8 year

Explanation:

GIVEN DATA:

the half life of the CO-60 is, T_1/2 = 5.27 years = 1.663 e+8 s

activity dN/dt = 1 mCi = 3.7 X 10^7 decay/s

activity ,  dN/dt = N* \lambda

              \frac{dN}{dt} = N* \frac{ln2}{T_1/2}

                   N = \frac{(dN/dt )(T1/2)}{ln2}

                      = ( 3.7 X 10^7 )(1.663*10^8 ) / ln2

                      = 8.877*10^{16}

Number of moles:

     n = N/NA = 8.877*10^{16} / 6.022X10^23 = 1.474*10^{-7} mol

mass of the CO-60 is,

   m = n*M = [1.474*10^{-7} mol]*[59.93 grams /mol] = 8.81*10^{-6}grams

-----------------------------------------------------------------------------------------

    time t = -[T1/2 / ln2]*ln[N/N0]

             = - [5.3 years / ln2]*ln[1x10-6/1x10-3]

             = 52.8 year

8 0
3 years ago
One of two 25-year-old identical twins begins a trip on a spaceship traveling at 0.8 c while her twin remains on Earth. The twin
borishaifa [10]

Answer:

This equation is based on twin paradox - a phenomena where one of the twin travels to space at a speed close to speed of light and the other remains on earth. the twin from the space on return discovers that the one on earth age faster.

Solution:

t_{o} = 10 years

v = 0.8c

c = speed of light in vacuum

The problem can be solved by time dilation equation:

t = \frac{t_{o}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}            (1)

where,

t = time observed from a different inertial frame

Now, using eqn (1), we get:

t = \frac{10}{\sqrt{1 - \frac{(0.8c)^{2}}{c^{2}}}}

t = 16.67 years

The age of the twin on spaceship according to the one on earth = 25+16.67 =41.66 years

8 0
3 years ago
Which of these stopped operating in 2011
Vladimir [108]
D. space shuttle program
8 0
3 years ago
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