Ask question

Ask question

All categories

2 years ago

13

A 75.0 kg man sits on a massless cart that is on a horizontal surface. The cart is initially stationary and it can move without

friction or air resistance. The man throws a 5.00 kg stone in the positive direction, applying a force to it so that it has acceleration +3.50 m/s2 as measured by a nearby observer on the ground. What is the man's acceleration during the throw, as seen by the same observer? Be careful to use correct signs.

1 answer:

Anna11 [10]2 years ago

6 0

Answer: $0.233 m/s^2$

Explanation:

Given

Mass of man m=75 kg

mass of stone $m_0=5 kg$

acceleration of stone $a=3.5 m/s^2$

To provide a acceleration of 3.5 m/s^2[/tex] he must exert a force of

$F=m_0\times a=5\times 3.5=17.5 N$

same Force is exerted by stone on cart

$17.5=m\times a$

$a=\frac{17.5}{75}=0.233 m/s^2$

You might be interested in

Find the area of the part of the plane 5x 3y z = 15 that lies in the first octant.

Lunna [17]

This part of the plane lies above a triangle with boundaries $x=0$ and $y=0$ along the coordinate axes, as well as the line

$z=0 \implies 5x + 3y = 15 \implies y = \dfrac{15 - 5x}3$

When $y=0$ , we have $15-5x=0\implies x=3$ . So this triangle is the set

$T = \left\{(x,y)\in\Bbb R^2 ~:~ 0\le x\le3 \text{ and } 0\le y\le\dfrac{15-5x}3\right\}$

Also, when $x=0$ , we have $y=\frac{15}3=5$ . So the triangle has length 3 and width 5, hence area 1/2•3•5 = 15/2.

Let $z=f(x,y) = 15 - 5x - 3y$ . Then the area of the plane over $T$ is

$\displaystyle \iint_T dA = \iint_T \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dx \, dy$

We have

$\dfrac{\partial f}{\partial x} = -5 \implies \left(\dfrac{\partial f}{\partial x}\right)^2 = 25$

$\dfrac{\partial f}{\partial y} = -3 \implies \left(\dfrac{\partial f}{\partial y}\right)^2 = 9$

$\implies\displaystyle \iint_T dA = \sqrt{35} \iint_T dx\,dy = \boxed{\frac{15\sqrt{35}}2}$

since the integral

$\displaystyle \iint_TdA$

is exactly the area of $T$ , 15/2.

8 0

1 year ago

An artillery shell is fired at an angle to the horizontal. Its initial velocity has a vertical component of 150 meters per secon

Nana76 [90]

Answer:

300.16m/s

Explanation:

An artillery shell is fired at an angle to the horizontal. Its initial velocity has a vertical component of 150 meters per second and a horizontal component of 260 meters per second. What is the magnitude of the initial velocity of the shell?

note velocity is the change in displacement to time

to find the magnitude of the initial speed ,

we will find the square of the vertical component plus the square of the horizontal component. then we look for the square root

U= $\sqrt{Ux^2+Uy^2}$

U= $\sqrt{260^2+150^2}$

U=90,100^2

U=300.16m/s

means that it covers 300.16m in 1 seconds

5 0

2 years ago

Which of the following leads to new discoveries and new theories?

Fynjy0 [20]

)scientific knowledge

3 0

2 years ago

Read 2 more answers

List two of the number of strategies you have learned

jolli1 [7]

Answer:

what?

Explanation:

does that means also i learned Ratios and multiplication i keared them like 5 years ago?

7 0

1 year ago

When energy is transferred to air, what happens to the particles of air? (1 point) They move slower. They move faster. They cool

zubka84 [21]

This question is a critical question. as we all know, when energy is added to any state of water, the particles move faster. and when energy is taken away from any state of water, the particles reduce speed. same with the particles of air. when energy is added; they move faster. when energy is removed; they move slower. so the answer is they move faster

5 0

3 years ago

Read 2 more answers