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3 years ago

10

What force is needed to accelerate a 90 kg punching bag 13 m/s2?

1 answer:

bagirrra123 [75]3 years ago

7 0

Answer:

The answer is 1170 Newton

Explanation:

Force =. Mass × Acceleration

= 90 × 13

= 1170 Newton

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A worker on the roof of a house drops his 0.58 kg hammer, which slides down the roof at constant speed of 6.69 m/s. The roof mak

shusha [124]

Answer:

17.3 m

Explanation:

Given that,

Mass of a hammer is 0.58 kg

Velocity with which the hammer slides is 6.69 m/s at constant speed.

The roof makes an angle of 18 ◦ with the horizontal, and its lowest point is 18.2 m from the ground. We need to find the horizontal distance traveled by the hammer between the time is leaves the roof of the house and the time it hits the ground. Firstly, we will find the time taken by the hammer when it reaches ground in vertical direction.

$y=y_0+v_ot +\dfrac{1}{2}gt^2$

Putting all the values,

$0=18.2+6.69t-\dfrac{1}{2}\times 9.8t^2\\\\-4.9t^2+6.69t+18.2=0\\\\t=\dfrac{-6.69+\sqrt{6.69^{2}-4\cdot-4.9\cdot18.2}}{2\cdot-4.9}, \dfrac{-6.69-\sqrt{6.69^{2}-4\cdot-4.9\cdot18.2}}{2\cdot-4.9}\\\\t=-1.36\ s\ \text{and}\ t=2.72\ s$

Neglecting negative value,

To find horizontal distance, multiply 2.72 s with the horizontal component of velocity.

$d=2.72\times 6.69\times \cos(18)\\\\d=17.3\ m$

3 0

4 years ago

A hollow sphere of inner radius 8.82 cm and outer radius 9.91 cm floats half-submerged in a liquid of density 948.00 kg/m^3. (a)

kari74 [83]

Answer:

a) 0.568 kg

b) 474 kg/m³

Explanation:

Given:

Inner radius = 8.82 cm = 0.0882 m

Outer radius = 9.91 cm = 0.0991 m

Density of the liquid = 948.00 Kg/m³

a) The volume of the sphere = $\frac{4\pi}{3}\times(0.0991^2-0.0882^2)$

or

volume of sphere = 0.0012 m³

also, volume of half sphere = $\frac{\textup{Total volume}}{\textup{2}}$

or

volume of half sphere = $\frac{\textup{0.0012}}{\textup{2}}$

or

Volume of half sphere =0.0006 m³

Now, from the Archimedes principle

Mass of the sphere = Weight of the volume of object submerged

or

Mass of the sphere = 0.0006× 948.00 = 0.568 kg

b) Now, density = $\frac{\textup{Mass}}{\textup{Volume}}$

or

Density = $\frac{\textup{0.568}}{\textup{0.0012}}$

or

Density = 474 kg/m³

8 0

3 years ago

Inertia is the resistance of an object to a change in its....

mixas84 [53]

Inertia is the resistance of an object to a change in its B. Motion

4 0

3 years ago

Read 2 more answers

A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra

blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

$y=\sqrt{1+x^3}$

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Now at (2,3)

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s$

7 0

3 years ago

How does the value of gravity vary as you go:

iragen [17]

Answer:

Gravity changes with altitude. as we know The gravitational force is proportional to 1/R2, where R is your distance from the center of the Earth.

eg. The radius of the Earth at the equator is 6400 kilometers.

Let's say you were in a jet at the equator that was 40 kilometers high above the earth's surface.

may be helpfull

6 0

2 years ago