All categories

2 years ago

5. Emanuel Zacchini, the famous human cannonball of the Ringling Bros. and Barnum and Bailey Circus,

Physics

1 answer:

Strike441 [17]2 years ago

3 0

Explanation:

सिद्ध कीजिए किसी भी बराबर भुजाओं वाले त्रिभुज में उनके सामने के कोण बराबर होते है

You might be interested in

How many planets are there?

Y_Kistochka [10]

Answer:

There are eight planets in our Solar System.

Explanation:

4 0

3 years ago

Read 2 more answers

How far can your little brother get if he can travel at 2.5 m/s and in 5?

KatRina [158]

d=? v=2.5 u=0 and t=5 therefore the formula to be used to find the distance my brother covered is d=1/2(v-u)t

d=1/2(2.5-0)5

=6.15m

4 0

2 years ago

Suppose that a block of mass 2 kg is pulled to the right with a force of 10 N, and the friction force on the block is directed t

Law Incorporation [45]

Answer:

The block has an acceleration of $3 m/s^{2}$

Explanation:

By means of Newton's second law it can be determine the acceleration of the block.

$\sum F_{r} = ma$ (1)

Where $\sum F_{r}$ represents the net force, m is the mass and a is the acceleration.

$F_{x} + F{y} = ma$ (2)

The forces present in x are $F = 10 N$ and $f = 4 N$ (the friction force):

$F_{x} = 10 N - 4 N$

Notice that $f$ subtracts to $F$ since it is at the opposite direction.

$F_{x} = 6 N$

The forces present in y balance each other:

$F_{y} = 0$

Therefore:

$6 + 0 = ma$

$6 N = (2kg)a$ (3)

But $1 N = 1 Kg.m/s^{2}$ and writing (3) in terms of a it is get:

$a = \frac{6 Kg.m/s^{2}}{2 Kg}$

$a = 3 m/s^{2}$

So the block has an acceleration of $a = 3 m/s^{2}$ .

4 0

3 years ago

what velocity must a 1340kg car have in order to havw the same momentum as a 2680 kg truck traveling at a velocity of 15m/s to t

kykrilka [37]

Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.

Hope i helped
Have a good day :)

6 0

3 years ago

A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$

The box is initially at rest, so $v_i=0$ . Solving for $v_f$ :

$v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}$

5 0

3 years ago