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3 years ago

Why does a third class lever cannot magnify force?

Physics

1 answer:

maw [93]3 years ago

5 0

Explanation:

The third class lever cannot magnify our force because in third class lever the effort it between the load and the fulcrum. Also, in this type of lever no matter where the force is applied, it is always greater than the force of load. Hence, That type of lever cannot magnify our force.

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2. What current flows through a hair dryer plugged into a 110 Volt circuit et it as a te

JulsSmile [24]

Explanation:

V =I*R

I= V/I

I=110/25

I= 4.4 Amps

8 0

3 years ago

5.00 kg of liquid water is heated to 100.0 °C in a closed system. At this temperature, the density of liquid water is 958 kg/m3.

user100 [1]

Answer:

$1.04\times 10^7\ J.$

Explanation:

In the question given :

Pressure is constant

Therefore, Work done, $W=P\times\Delta V$

Pressure, P=1.01 × 105 Pa.

Final volume, $V_f=8.50\ m^3.$

Initial volume, $V_i=\dfrac{Mass}{density}=\dfrac{5}{958}=5.22\times10^-3\ m^3.$

Therefore, W=8.58\times 10^{5}\ J.

Also, Heat Given, $Q=m\times L=5\times 2.26\times 10^{6}\ J=1.13\times 10^7\ J.$

Also, according to First law of thermodynamics:

$\Delta U=Q-W=(1.13\times 10^7)-(8.58\times 10^5)=1.04\times 10^7\ J.$

Hence, this is the required solution.

8 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

Time taken to reach 1180 m is 11.29 seconds

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

Time the rocket will keep going up after the engines shut off is 13.06 seconds.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$