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kupik [55]
3 years ago
14

A 50kg block is at rest on the ground.

Physics
1 answer:
yKpoI14uk [10]3 years ago
7 0

Answer:

(b) 490 N downwards

Explanation:

(b) Formula : w= mg

m is the mass of the object

g is a gravitational constant (9.8)

w= mg

= 50×9.8

= 490 N downwards

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It’s burning wax. All of the other options are physical changes .
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Which graph accurately shows the relationship between kinetic energy and mass as mass increases
gtnhenbr [62]

Answer:

c

Explanation:

energy doesnt affect to mass of a object

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Identify the independent, dependent, and constant variables for different experiments.
diamong [38]

ANSWER:

IV, Type of dish detergent. DV, height of foam. CV, type of container, amount of water in container, temperature of water, time the container is agitated.

Explanation:

Independent variable(IV)- what you change during the experiment.

dependent variable(DV)- what you're measuring during an experiment. The dependent variable is DEPENDENT because it's results DEPEND on the independent variable at play.

Constant variables(CV)- things that do not change in order to isolate the tested variables as much as possible.

3 0
3 years ago
A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s
Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is:   w_{c} =m_{c} g   and maximum tension is T=2.50 m_{c} g=1.41*10^4N

total mass and weight is :

M =m_{c}+ m_{b} =740kg+550kg=1290 kg

w_{M} =1.2650*10^4N

∑F_{y} =ma_{y}

T-M_{g} =Ma_{y}

a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg

=0.645m/s^2

B)

maximum acceleration

a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0

using y-y_{0} =v_{oy} t+1/2(a_{y} )t^2

to solve for t

t=\sqrt{2(y-y_{0} )/a_{y} }

t=\sqrt{2(119m)/0.645m/s^2} =19.20s

6 0
3 years ago
A scientist wants to model an internal organ with connective tissue as a mass on a spring. The mass of the organ is 2.0 kg, and
vichka [17]

Answer:

Spring constant, k = 5483.11 N/m

Explanation:

It is given that,

Mass of the organ, m = 2 kg

The natural period of oscillation is, T = 0.12 s

Let k is the spring constant for the spring in the scientist's model. The period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

k=\dfrac{4\pi^2 m}{T^2}

k=\dfrac{4\pi^2 \times 2\ kg}{(0.12\ s)^2}

k = 5483.11 N/m

So, the  spring constant for the spring in the scientist's model is 5483.11 N/m.

5 0
3 years ago
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