The acceleration of gravity is approximately 10 m/s2. What would the force of gravity be if the mass of an object is 4.5 kg?

Qualitative observation requires numerous data to discribe research A .true B. False

saveliy_v [14]

Answer:

true

Explanation:

Qualitative Observation is the research process of using subjective methodologies to gather information or data. Since the focus on qualitative observation is to equate quality differences, it is a lot more time consuming than quantitative observation but the sample size used is much smaller and the research is extensive and a lot more personal.

8 0

3 years ago

A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$

$V(t) = 38 \frac{m}{s} - g * t$

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

$0 m = 38 \frac{m}{s} - g * t$

and obtain:

$t = 38 \frac{m}{s} / g$

$t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}$

$t = 3.9 s$

Plugin this time on our first equation we find:

$y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2$

$y=73.67 m$

6 0

4 years ago

The tip of a nail exerts tremendous pressure when hit by a hammer because it exerts a large force over a small area. What force

yanalaym [24]

Answer:

Force, F = 2714.33 N

Explanation:

It is given that,

Diameter of the circular tip, d = 1.2 mm = 0.0012 m

Radius, r = 0.0006 m

Pressure acting on the hammer, $P=2.4\times 10^9\ N/m^2$

Let F is the force exerted on a nail. It can be calculated using the definition of pressure exerted. Its formula is given by :

$P=\dfrac{F}A}$

$F=P\times \pi r^2$

$F=2.4\times 10^9\ N/m^2 \times \pi (0.0006\ m)^2$

F = 2714.33 N

So, the force exerted on a nail 2714.33 N. Hence, this is the required solution.

4 0

4 years ago

A solid steel bar of circular cross section has diameter d 5 2.5 in., L 5 60 in., and shear modulus of elasticity G 5 11.5 3 106

8090 [49]

Answer:

A) θ = 4.9 x 10^(-3) rad

B) τ_max = 1.173 ksi

C) τ_a = 4.786 ksi

Explanation:

We are given;

diameter; d = 2.5 inches = 0.2083 ft

Length; L = 60 inches = 5 ft

Torque; T = 300 lb.ft

Shear modulus; G = 11.5 x 10^(6) psi = 11.5 x 144 x 10^(6) lb/ft² = 1.656 x 10^(9) lb/ft²

A) Now, formula to determine angle of twist is given as;

T/I_p = Gθ/L

Where I_p is polar moment of inertia

θ is angle of twist.

Now I_p = πd⁴/32 = π(0.2083)⁴/32 = 1.85 x 10^(-4) ft⁴

Thus, making θ the subject, we have;

TL/GI_p = θ

θ = (300 x 5)/(1.656 x 10^(9) x 1.85 x 10^(-4))

θ = 4.9 x 10^(-3) rad

B) Maximum shear stress is given by the formula ;

τ_max = (Gθ/L)(d/2)

From earlier, (Gθ/L) = T/I_p

Thus, (Gθ/L) = 300/1.85 x 10^(-4) = 1621621.6216

Thus,

τ_max = 1621621.6216 x (0.2083/2)

τ_max = 168891.89 lbf/ft²

Converting to ksi = 168891.89/144000 ksi = 1.173 ksi

C) Shear stress at radial distance is given as;

τ_a = (Gθ/L)•r_a

r_a is given as 5.1 inches = 0.425m

τ_a = 1621621.6216 x 0.425 = 689189.189 lbf/ft²

Converting to ksi = 689189.189/144000 ksi = 4.786 ksi

7 0

3 years ago

a rock is vertically upward with a velocity of 10 m/s. calculate the maximum height it reaches and time taken to reach that heig

lina2011 [118]

Answer:

maximum height: p(t) = Vo * t - 1/2 * g * t^2

p’(t) = v(t) = 0 = Vo - g*t. So, maximum height occurs when t = Vo / g

p(Vo / g) = Vo^2/g - 1/2 * g * (Vo/g)^2

Vo = 10 m / s. Let’s approximate g = 10 m / s^2

p(Vo / g) = 10^2 / 10 - 1/2 * 10 * (10/10)^2 = 10 - 5 = 5 meters (approximately)

Calculation of time:

v = u + gt

0 = 10√2 + (-10)t

-10√2 = -10t

2 = √2s

8 0

3 years ago