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3 years ago

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a shopping bag can provide an upward force of 65N before breaking. A shopper puts 5 kg of groceries in the bag. If the shopper t

ries to lift the bag with an upward acceleration of 2 m/s/s, will the bag break?

1 answer:

kipiarov [429]3 years ago

6 0

Answer:

Since Force is Mass * Acceleration, the equation would be:

5kg * 2m^2 = 10N

<em>(10N is less than 65N)</em>

Therefore, it would not be enough force for the bag to break.

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1. Is the relationship between velocity and centripetal force a direct, linear relationship or is it a nonlinear square relation

Svetllana [295]

Answer:

non linear square relationship

Explanation:

formula for centripetal force is given as

a = mv^2/r

here a ic centripetal acceleration , m is mass of body moving in circle of radius r and v is velocity of body . If m ,and r are constant we have

a = constant × v^2

a α v^2

hence non linear square relationship

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3 years ago

If a 15 kg mass accelerates at a rate of 4 m/s2, what net force acts on it?

jarptica [38.1K]

<h2>Answer: D 60N</h2>

<h3>Explanation:</h3>

Mass(M)=15 kg

Acceleration(A)=4 m/s2

Force=?

Now,

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F=15×4

F=60N Ans

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3 years ago

A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.Part AWhat is the

Mademuasel [1]

Answer:

(a) $a=2m/sec^2$

(b) 5220 j

(c) 1740 watt

(d) 3446.66 watt

Explanation:

We have given mass m = 290 kg

Initial velocity u = 0 m/sec

Final velocity v = 6 m/sec

Time t = 3 sec

From first equation of motion

v = u+at

So $a=\frac{v-u}{t}=\frac{6-0}{3}=2m/sec^2$

(a) We know that force is given by

F = ma

So force will be $F=290\times 2=580N$

(b) From second equation of motion we know that

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We know that work done is given by

W = F s = 580×9 =5220 j

(c) Time is given as t = 3 sec

We know that power is given as

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(d) Time t = 1.5 sec

So $P=\frac{W}{t}=\frac{5220}{1.5}=3466.66Watt$

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4 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

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kykrilka [37]

Answer:

The second one, the higher the hertz the higher the frequency we hear

Explanation:

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3 years ago