If a bowling ball rolls 12 meters in 2.2 seconds what is the speed ?

Two boxes are connected to each other by a string as shown in the figure. The 10-n box slides without friction on the horizontal

FromTheMoon [43]

Answer:

T=7.4 N hence T<30 N

Explanation:

The figure is likely to be similar to the one attached. Writing the equation for forces we have

F-T=Fa/g where F is the force, T is tension, a is acceleration and g is acceleration due to gravity. Substituting the figures we have the first equation as

30 N - T = (30/9.81)a

Also, we know that T=F*a/g and substituting 10N for F we obtain the second equation as

T = (10/9.81)a

Adding the first and second equations we obtain

30 = 4.077471967

a Hence

$a=\frac {30}{4.077471967}=7.3575 m/s^{2}$

and T=a hence

T is approximately 7.4 N

5 0

3 years ago

Read 2 more answers

What is the type of intermolecular force in methylated spirits

Olin [163]

. Methylated spirits have ethanol as a base but may include methyl alcohol (methanol) as part of the denaturing process.

7 0

3 years ago

A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0Â° with the horizontal. The coeff

natulia [17]

Answer:

a

$H =212.6 \ J$

b

$v = 7.647 \ m/s$

Explanation:

From the question we are told that

The child's weight is $W_c = 287 \ N$

The length of the sliding surface of the playground is $L = 7.20 \ m$

The coefficient of friction is $\mu = 0.120$

The angle is $\theta = 31.0 ^o$

The initial speed is $u = 0.559 \ m/s$

Generally the normal force acting on the child is mathematically represented as

=> $N = mg * cos \theta$

Note $m * g = W_c$

Generally the frictional force between the slide and the child is

$F_f = \mu * mg * cos \theta$

Generally the resultant force acting on the child due to her weight and the frictional force is mathematically represented as

$F =m* g sin(\theta) - F_f$

Here F is the resultant force and it is represented as $F = ma$

=> $ma = m* g sin(31.0) - \mu * mg * cos (31.0)$

=> $a = g sin(31.0)- \mu * g * cos (31.0)$

=> $a = 9.8 * sin(31.0) - 0.120 * 9.8 * cos (31.0)$

=> $a = 4.039 \ m/s^2$

So

$F_f = 0.120 * 287 * cos (31.0)$

=> $F_f = 29.52 \ N$

Generally the heat energy generated by the frictional force which equivalent tot the workdone by the frictional force is mathematically represented as

$H = F_f * L$