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3 years ago

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A ball was thrown vertically upward on a planet with a velocity of 2.5 m/sec. If it can rise up to 4m, then calculate gravitatio

nal strength of the planet.
please help ill give brainlest

1 answer:

svlad2 [7]3 years ago

5 0

Answer:

sorry not today Yazan was here

Explanation:

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Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,

klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble= $2\hat{i}-3\hat{k}$ N

Position vector= $r=0.5\hat{j}-2\hat{k}$ m

a.We have to find the resulting torque on the pebble about origin.

Torque= $r\times F$

Substitute the values then we get

$Torque= (0.5j-2k)\times (2i-3k)$

Torque= $-k-1.5i-4j$ N-m

By using $i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0$

b. $r=2i-3k$

$r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k$

Torque about point (2,0,-3)

$\tau=(-2i+0.5j+k)\times (2i-3k)$

$\tau=-6j-k-1.5i+2j=-1.5i-4j-k$ N-m

6 0

3 years ago

LenKa [72]

Mu wildest guess is c 1500

7 0

4 years ago

Read 2 more answers

What happens when infrared hits a shiny surface

saw5 [17]

Will reflect ........

3 0

4 years ago

a golfer hits a 0.05 kg golf ball with an impulse of 3 N-s what is the change in velocity of the ball

Rainbow [258]

so your saying the start is 0 N and when he/she hits the ball its inertia is 3 N. if that is so m*v=

.05*3=<u>.15</u>

7 0

4 years ago

A wire with a linear mass density of 1.17 g/cm moves at a constant speed on a horizontal surface and the coefficient of kinetic

stira [4]

Answer:

The value is $B = 0.2312 \ T$

The direction is into the surface

Explanation:

From the question we are told that

The mass density is $\mu =\frac{m}{L} = 1.17 \ g/cm =0.117 kg/m$

The coefficient of kinetic friction is $\mu_k = 0.250$

The current the wire carries is $I = 1.24 \ A$

Generally the magnetic force acting on the wire is mathematically represented as

$F_F = F_B$

Here $F_F$ is the frictional force which is mathematically represented as

$F_F = \mu_k * m * g$

While $F_B$ is the magnetic force which is mathematically represented as

$F_B = BILsin(\theta )$

Here $\theta =90^o$ is the angle between the direction of the force and that of the current

So

$F_B = BIL$

So

$BIL = \mu_k * m * g$

=> $B = \mu_k * \frac{m}{L} * [\frac{g}{I} ]$

=> $B = 0.25 * 0.117 * [\frac{9.8}{1.24} ]$

=> $B = 0.2312 \ T$

Apply the right hand curling rule , the thumb pointing towards that direction of the current we see that the direction of the magnetic field is into the surface as shown on the first uploaded image

8 0

3 years ago