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3 years ago

What is projectile motion.????

Engineering

2 answers:

True [87]3 years ago

8 0

Answer:

Projectile motion is the motion of a body which experiences both vertical and horizontal motions ( trajection ) from point of flight up to the point of landing.

WARRIOR [948]3 years ago

7 0

Answer:

Projectile motion is a form of motion experienced by a launched object.

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For a metal that has a yield strength of 690 MPa and a plane strain fracture toughness (KIc) of 32 MPa-m1/2, compute the minimum

Digiron [165]

Answer:

the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm

Explanation:

Given the data in the question;

yield strength σ $_y$ = 690 Mpa

plane strain fracture toughness K $_{Ic$ = 32 MPa- $m^{1/2$

minimum component thickness for which the condition of plane strain is valid = ?

Now, for plane strain conditions, the minimum thickness required is expressed as;

t ≥ 2.5( K $_{Ic$ / σ $_y$ )²

so we substitute our values into the formula

t ≥ 2.5( 32 / 690 )²

t ≥ 2.5( 0.0463768 )²

t ≥ 2.5 × 0.0021508

t ≥ 0.005377 m or 5.38 mm

Therefore, the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm

7 0

3 years ago

What type of fire extinguisher is used for electrical fires.

AysviL [449]

The all purpose one will work or you could use type C I believe.

8 0

3 years ago

Air initially at 15 psla and 60 F is compressed to 75 psia and 400 F. The power input to air under steady state condition is 5 h

Triss [41]

Answer: $\dot{m}=3.46lbm/min$

Explanation:

Initial conditions

$P_1=15 psia$

$T_1=60 F^{\circ}$

Final conditions

$P_2=75 psia$

$T_2=400F^{\circ}$

Steady flow energy equation

$\dot{m}\left [ h_1+\frac{v_1^2}{2}+gz_1\right ]+\dot{Q}=\dot{m}\left [ h_2+[tex]\frac{v_2^2}{2}+gz_2\right ]+\dot{W}$

$\dot{m}\left [ c_pT_1+\frac{0^2}{2}+g0\right ]+\dot{Q}=\dot{m}\left [ c_pT_2+\frac{0^2}{2}+g0\right ]+\dot{W}$

$\dot{m}c_p\left [ T_1-T_2\right ]+\left [ -5hp\right ]=\dot{W} -5\times 746\times 3.4121$

$-4\dot{m}-\dot{m}\times 0.24\times \left [ 400-60\right ]$

$-81.6\dot{m}-4\dot{m}=-4.949 BTU/sec$

$\dot{m}=0.057821lbm/sec$

$\dot{m}=3.46lbm/min$

3 0

3 years ago

Refrigerant-134a enters an adiabatic compressor as saturated vapor at -24°C and leaves at 0.8 MPa and 60°C. The mass flow rate o

Leni [432]

Answer:

(a) The power input to the compressor: $\dot{W}=73.07 kJ/s = 73.07 kW$

(b) The volume flow rate of the refrigerant at the compressor inlet: $\dot{v}=0.209 m^{3}/s$

Explanation:

(a)

We need to check the values of enthalpy (as we have an open system) for both states, being the inlet, state 1 and the outlet, state 2. We will know these values by checking the vapor charts of R134a, I used the ones found in Thermodynamics of Cengel, 7th edition.

Then, our values are:

$h_{1}=235.92kJ/kg\\h_{2}=296.81kJkg$

Now we proceed to know the work with the following expression:

$\dot{W}=\dot{m}(h_{2}-h_{1})$

Now we replace values and our result is:

$\dot{W}=73.07 kJ/s = 73.07 kW$

(b)

To know the volume rate at the compressor inlet, we need to know the specific volume in that phase, as we have that is saturated and at -24°C, we can read our table:

$\nu=0.1739m^{3}/kg$

With our specific volume and the mass rate, we can calculate the volume rate:

$\dot{v}=\nu * \dot{m}\\\dot{v}=0.209 m^{3}/s$

6 0

3 years ago

Three bars each made of different materials are connected together and placed between two walls when the temperature is 12 oC. D

slega [8]

Answer:

F = 9.11 x 10³ N = 9.11 KN

Explanation:

The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:

ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper

ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)

where,

ΔL = Equivalent Change in Length = ?

ΔT = Change in Temperature = 25°C - 12°C = 13°C

Ls = Length of Steel Segment = 300 mm = 0.3 m

Lb = Length of Brass Segment = 200 mm = 0.2 m

Lc = Length of Copper Segment = 100 mm = 0.1 m

Therefore,

ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)

ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m

ΔL = 123.5 x 10⁻⁶ m ----------------------- equation (1)

Now, we calculate this deflection in terms of an applied force (F):

ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)

ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)

ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)

ΔL = F(13.55 x 10⁻⁹ m/N) --------------------- equation (1)

Comparing equation (1) and equation (2):

123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)

F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)

6 0

3 years ago

What is projectile motion.????​

What is projectile motion.????