Putting a marshmallow directly in the fire

Salt Flats of Utah, had a mass of 4100 kg, and its engine

Andrej [43]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{68000}{4100} = \frac{680}{41} \\ = 16.58536$

We have the final answer as

Hope this helps you

8 0

3 years ago

A 1250-kg compact car is moving with velocity v1 =36.2i^+12.7j^m/s. It skids on a frictionless icy patch and collides with a 448

MA_775_DIABLO [31]

Momentum is conserved, so the sum of the separate momenta of the car and wagon is equal to the momentum of the combined system:

(1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s) = ((1250 + 448) kg) <em>v</em>

where <em>v</em> is the velocity of the system. Solve for <em>v</em> :

<em>v</em> = ((1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s)) / (1698 kg)

<em>v</em> ≈ (30.3 <em>i</em> + 12.0 <em>j</em> ) m/s

7 0

3 years ago

Supposing d(t) is known to have value D,

creativ13 [48]

Answer:

The procedure is: solve the quadratic equation for $t$ .

Explanation:

This question assumes uniformly accelerated motion, for which the distance d a particle travels in time t is given by the general equation:

$d(t)=d_0+v_0t+at^2/2$

That is a quadratic equation, where the independent variable is the time $t$ .

Thus, the procedure that will find the time t at which the distance value is known to be D is to solve the quadratic equation for $t$ .

To solve it you start by changing the equation to the general form of the quadratic equations, rearranging the terms:

$(a/2)t^2+v_0t+(d_0-D)=0$

Some times that equation may be solved by factoring, and always it can be solved by using the quadratic formula:

$t=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

Where:

$a=-a/2\\ \\ b=v_0\\ \\ c=d_0-D$

That may have two solutions. Some times one of the solution makes no physical sense (for example time cannot be negative) but others the two solutions are valid.

5 0

3 years ago

A person walks into a room that has, on opposite walls, two plane mirrors producing multiple images. Find the distances from the

Lapatulllka [165]

Answer:13.2,41.4,82.8 ft

Explanation:

Given Person is 6.60 ft from the left hand side mirror

Since the focal length of plane mirror is infinity therefore image and object are equidistant from mirror

Distance of first image on left mirror is

$d_1=6.6+6.6=13.2 ft$

i.e. image is 13.2 ft away from object

Second image

Now the right mirror forms the image of object at distance of 14.1 ft right from right mirror so its image is formed in left mirror at a distance of 34.8 from left mirror

so image distance from person is

$d_2=34.8+6.6=41.4 ft$

Third image now form image of second image is formed on right mirror at a distance of 55.5 right from right mirror

and its mirror image is formed on left mirror at a distance of 76.2 left

from left mirror

$d_3=76.2+6.6=82.8 ft$

8 0

3 years ago

3. How much work is done when you pull a 6 N wagon for 5 meters?

8090 [49]

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question

force = 6 N

distance = 5 m

We have

workdone = 6 × 5 = 30

We have the final answer as

Hope this helps you

3 0

3 years ago