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3 years ago

In what direction or orientation did the man throw the ball in image A? image B?

Physics

1 answer:

Ray Of Light [21]3 years ago

4 0

Answer:

dear where is image

how can we answer you

without image

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In the picture below, a doctor is having a video conference with a patient. What might be a positive effect of the technology be

GarryVolchara [31]

Answer:

Sorry to say but where is the photo???

The positive effect of technology being used might be using sethescope or checking BP rate if it is good.

Explanation:

6 0

3 years ago

Now imagine a person dragging a 50 kg box along the ground with a rope, as

ANTONII [103]

Answer:

The coefficient of static friction between the box and floor is, μ = 0.061

Explanation:

Given data,

The mass of the box, m = 50 kg

The force exerted by the person, F = 50 N

The time period of motion, t = 10 s

The frictional force acting on the box, f = 30 N

The normal force on the box, η = mg

= 50 x 9.8

= 490 N

The coefficient of friction,

μ = f/ η

= 30 / 490

= 0.061

Hence, the coefficient of static friction between the box and floor is, μ = 0.061

7 0

4 years ago

How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab

tigry1 [53]

Answer:

Vy = 26 m/s sin 30 = 13 m/s vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec time to reach Vy = 0

H = Vy t + 1/2 g t^2

H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

4 0

1 year ago

A block of mass M on a horizontal surface is connected to the end of a massless spring of spring constant k. The block is pulled

slamgirl [31]

Answer:

Minimum coefficient of kinetic friction between the surface and the block is $\mu_k=\frac{kx}{2Mg}$ .

Explanation:

Given:

Mass of the block = M

Spring constant = k

Distance pulled = x

According to the question:

We have to find the minimum co-efficient of kinetic friction between the surface and the block that will prevent the block from returning to its equilibrium with non-zero speed.

So,

From the FBD we can say that:

⇒ Normal force, $N=Mg$ ...equation(i)

⇒ Elastic potential energy, $PE$ = $\frac{kx^2}{2}$ ...equation (ii)

⇒ Frictional force, $f$ = $\mu_kN$ ...equation (iii)

⇒ Plugging (i) in (iii).

⇒ $f=\mu_kMg$

Now,

⇒ As we know that the energy lost due to friction is equivalent to PE .

⇒ $PE=fx$ ...considering PE as $mgh$ or $f(x)$ .

Arranging the equation.

⇒ $\frac{kx^2}{2}=\mu_k Mg (x)$

⇒ $\frac{kx}{2}=\mu_k Mg$ ...eliminating x from both sides.

⇒ $\frac{kx}{2Mg}=\mu_k$ ...dividing both sides wit Mg.

Minimum coefficient of kinetic friction between the surface and the block is $\frac{kx}{2Mg}=\mu_k$ .

4 0

4 years ago

A truck’s suspension spring each have a spring constant of 769 N/m. If the potential energy of the right front spring is 250 J,

Alex Ar [27]

Answer:

x = 0.81 m

Explanation:

given,

spring constant, k = 769 N/m

Potential energy of the spring = 250 J

distance of spring compression = ?

using conservation of energy

potential energy will equal to the spring energy

$PE = \dfrac{1}{2}kx^2$

$250= \dfrac{1}{2}\times 769\times x^2$

x² = 0.650

x = 0.81 m

Hence, the spring is compressed to 0.81 m

6 0

3 years ago

In what direction or orientation did the man throw the ball in image A? image B?​

In what direction or orientation did the man throw the ball in image A? image B?