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3 years ago

14

A boy is pulling a wagon with the force of 5N. A girl is pushing on the same wagon in the same direction with a force of 3N. Wha

t is the new force acting on this wagon?

1 answer:

frez [133]3 years ago

8 0

The new force that is acting on the wagon is 8N I believe

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A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

3 0

3 years ago

A 55-kg woman is wearing high heels.

Grace [21]

Answer:

Pressure, $P=1.90\times 10^7\ Pa$

Explanation:

It is given that,

Mass of the woman, m = 55 kg

Diameter of the circular cross section, d = 6 mm

Radius, r = 3 mm = 0.003 m

Let P is the pressure exerted on the floor. It is equal to the force acting on woman per unit area. It is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{mg}{\pi r^2}$

$P=\dfrac{55\times 9.8}{\pi (0.003)^2}$

$P=1.90\times 10^7\ Pa$

So, the pressure exerted on the floor is $1.90\times 10^7\ Pa$ . Hence, this is the required solution.

7 0

3 years ago

Delvig [45]

Answer:

Approximately $73\; {\rm N}$ , assuming that the acceleration of this ball is constant during the descent.

Explanation:

Assume that the acceleration of this ball, $a$ , is constant during the entire descent.

Let $x$ denote the displacement of this ball and let $t$ denote the duration of the descent. The SUVAT equation $x = (1/2)\, a\, t^{2}$ would apply.

Rearrange this equation to find an expression for the acceleration, $a$ , of this ball:

$\begin{aligned} a &= \frac{2\, x}{t^{2}}\end{aligned}$ .

Note that $x = 11\; {\rm m}$ and $t = 1.5\; {\rm s}$ in this question. Thus:

$\begin{aligned} a &= \frac{2\, x}{t^{2}} \\ &= \frac{2 \times 11\; {\rm m}}{(1.5\; {\rm s})^{2}} \\ &\approx 9.78\; {\rm m \cdot s^{-2}}\end{aligned}$ .

Let $m$ denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is $a$ , the net external force on this ball would be $m\, a$ .

Since $m = 7.5\; {\rm kg}$ and $a \approx 9.78\; {\rm m\cdot s^{-2}}$ , the net external force on this ball would be:

$\begin{aligned} (\text{net force}) &= m\, a \\ &\approx 7.5\; {\rm kg} \times 9.78\; {\rm m\cdot s^{-2}} \\ &\approx 73\; {\rm kg \cdot m \cdot s^{-2} \\ &= 73\; {\rm N} && (1\; {\rm N} = 1\; {\rm kg \cdot m\cdot s^{-2}}) \end{aligned}$ .

4 0

2 years ago

The power in an electrical circuit is given by the equation P = I2R, where I is the current flowing through the circuit and R is

kiruha [24]

Answer:

Explanation: Here we have given a direct equation . There for no need to worry .

P = I²×R

P = (12)² ×100

P = 14400W = 14.4 kW

For second one

P = I² ×R

200 = I²×150

I = √200/150

I = 1.15 A

5 0

2 years ago

The velocity of a car doubles over a short period of time. Which additional information is sufficient to determine the amount of

Rom4ik [11]

Answer:

Time and velocity

Explanation:

The time taken for the velocity to double is very important to find the amount of acceleration the car acquires.

Acceleration is the rate of change of velocity with time.

Acceleration = $\frac{v -u}{t}$

v is the initial velocity

u is the final velocity

t is the time taken

So, the velocity and time is needed to calculate the value of the acceleration the car undergoes.

4 0

3 years ago