How far moves in 95g ball by applying 200 Newton Force if 14 KJ of energy is transferred

What is the impulse when a test car traveling at 15 m/s strikes a cement wall at a force of 1,900,000 N if the impact lasts for

svp [43]

Answer:

impulse = force x time

impulse = 1900000 x 0.005 = 9500Ns

8 0

3 years ago

Given three vectors A = 24i + 33j, B = 55i - 12j and C = 2i + 43j (a) Find the magnitude of each vector. (b) Write an expression

slega [8]

Answer:

(a) , . and .

(b) $\vec A - \vec C=22 \hat i -10 \hat j$ .

(c) $|\vec A - \vec B|=63.13$ and the direction $\theta =$ 124.56°.

Explanation:

Given that,

,

and

$\vec {C}=2 \hat i +43 \hat j$

(a) The magnitude of a vector is the square root of the sum of the square of all the components of the vector, i.e. for a ,.

So, the magnitude of the is

$|\vec A|=\sqrt {24^2+ 33^2}$

$\Rightarrow |\vec A|=\sqrt {1665}$

.

The magnitude of the is

$|\vec B|=\sqrt {55^2+ (-12)^2}$

$\Rightarrow |\vec B|=\sqrt {3169}$

.

And, the magnitude of the is

$|\vec C|=\sqrt {2^2+ 43^2}$

$\Rightarrow |\vec C|=\sqrt {1853}$

.

(b) The difference between the two vectors is the difference between the corresponding components of the vectors. So, the required expression of is

$\vec A - \vec C=(24 \hat i +33 \hat j) - (2 \hat i +43 \hat j)$

$\Rightarrow \vec A - \vec C=24 \hat i +33 \hat j - 2 \hat i -43 \hat j$

$\Rightarrow \vec A - \vec C=22 \hat i -10 \hat j$

(c) The expression of is

$\vec A - \vec N=(24 \hat i +33 \hat j) - (55 \hat i -12 \hat j)$

$\Rightarrow \vec A - \vec B=24 \hat i +33 \hat j - 55\hat i +12 \hat j$

$\Rightarrow \vec A - \vec B=-31 \hat i +45 \hat j\;\cdots (i)$

The magnitude of is

$|\vec A - \vec B|=\sqrt {(-31)^2+55^2}$

$\Rightarrow |\vec A - \vec B|=\sqrt {3986}$

$\Rightarrow |\vec A - \vec B|=63.13$

Now, if a vector $\vec V= -\alpha \hat i +\beta \hat j$ in 3rd quadrant having direction $\theta$ with respect to $\hat i$ direction, than

in the anti-clockwise direction.

Here, from equation (i), for the vector $\vec A - \vec C$ , $\alpha=31$ and $\beta=45$ .

$\Rightarrow \theta = \pi-\tan ^{-1}\left(\frac {45}{31}\right)$

180°-55.44° [as \pi radian= 180°]

124.56° in the anti-clockwise direction.

(d) Vector diagrams for $\vec A +\vec B$ and $\vec A - \vec B$ has been shown

in the figure(b) and figure(c) recpectively.

Vector $\vec A - \vec B$ is in 3rd quadrant as calculated in part (c).

While Vector $\vec A +\vec B=(24 \hat i +33 \hat j)+(55 \hat i -12 \hat j)$

$\Rightarrow \vec A +\vec B=79 \hat i +21 \hat j$ , which is in 1st quadrant as both the components are position has been shown in figure(b).

6 0

3 years ago

The force of gravity on an object is constant all over the earth. True False

anastassius [24]

False.

Gravity changes depending on latitude, as well as height above sea level.

3 0

3 years ago

Read 2 more answers

An unknown mass is on a flat frictionless surface and a 2 kg object is attached to it with a string going over pulleyt mounted t

Lynna [10]

Answer:

Explanation:

F=ma

F in this case is the gravity acting on the 2kg object. Acceleration of gravity is 9.8 m/s^2. SF=ma, so

F = 2kg*(9.8 m/s^2) = 19.6 N

Now use this force to determine the mass of the object on the table:

F=ma

19.6 N (1N=kg*m/s^2) = m*(1.8 m/s^2)

m = 10.89 kg

3 0

3 years ago

Huck Finn walks at a speed of 0.70 m/sm/s across his raft (that is, he walks perpendicular to the raft's motion relative to the

exis [7]

Answer:

Explanation:

Given

Velocity of Huck w.r.t to raft $v_{H,raft}=0.7\ m/s$

Perpendicular to the motion of raft

Velocity of Raft in the river $v_{raft,river}=1.6\ m/s$

As Huck is traveling Perpendicular to the raft so he possess two velocities i.e. vertical velocity and horizontal velocity of River when observed from bank

$v_{Huck,river\ bank}=0.7\hat{j}+1.6\hat{i}$

So magnitude of velocity is given by

$|v|=\sqrt{0.7^2+1.6^2}$

$|v|=\sqrt{0.49+2.56}$

$|v|=\sqrt{3.05}$

$|v|=1.74\ m/s$

For direction $\tan =\frac{0.7}{1.6}=0.4375$

$\theta =23.63^{\circ}$ w.r.t river bank

4 0

3 years ago

How far moves in 95g ball by applying 200 Newton Force if 14 KJ of energy is transferred​

How far moves in 95g ball by applying 200 Newton Force if 14 KJ of energy is transferred